Mixing
Let c be an element of order 2 and d an element of order 5 in a Group G with 10 Elements , show that either...
$begingroup$
The first equality holds true if G is commutative, but I’m stuck on the second one if G isn’t. What’s a good way to start?
group-theory cyclic-groups
asked Jan 26 at 19:12
jk1998jk1998
264
$endgroup$
|
show 2 more comments
$begingroup$
The first equality holds true if G is commutative, but I’m stuck on the second one if G isn’t. What’s a good way to start?
group-theory cyclic-groups
asked Jan 26 at 19:12
jk1998jk1998
264
$endgroup$
2
$begingroup$
This is not true.
$endgroup$
– verret
Jan 26 at 19:21
1
$begingroup$
There is no reason $cdc^{-1}$ has to be a power of $d$. If for some reason you know that $cdc^{-1} = d^k$ for some $k$, then conjugating by $c$ twice gives $d = cd^kc^{-1} = (cdc^{-1})^k = (d^k)^k = d^{k^2}$, so $1 equiv k^2 bmod 5$ and thus $k equiv pm 1 bmod 5$. More generally, if $d$ has order $m$ and for some reason you know $cdc^{-1} = d^k$ then $k^2 equiv 1 bmod m$. Generally speaking there can be more solutions to $k^2 equiv 1 bmod m$ than just $pm 1 bmod m$, although those are the only solutions when $m$ is prime. The main problem is $cdc^{-1}$ need not be a power of $d$.
$endgroup$
– KCd
Jan 26 at 19:27
$begingroup$
"What's a good way to start?". A good way to start would be to check again the source of the problem and to write an exact formulation of the problem. Reading the title I have the impression that you wrote this down out of your memory, and some things are missing.
$endgroup$
– Dietrich Burde
Jan 26 at 19:37
$begingroup$
I’m sorry, I forgot that G needs to be a group of 10 elements
$endgroup$
– jk1998
Jan 26 at 19:41
$begingroup$
Hint: What are the possible groups with $10$ elements (up to isomorphism)?
$endgroup$
– Somos
Jan 26 at 19:42
|
show 2 more comments
$begingroup$
The first equality holds true if G is commutative, but I’m stuck on the second one if G isn’t. What’s a good way to start?
group-theory cyclic-groups
asked Jan 26 at 19:12
jk1998jk1998
264
$endgroup$
The first equality holds true if G is commutative, but I’m stuck on the second one if G isn’t. What’s a good way to start?
group-theory cyclic-groups
group-theory cyclic-groups
asked Jan 26 at 19:12
jk1998jk1998
264
asked Jan 26 at 19:12
jk1998jk1998
264
edited Jan 26 at 19:40
jk1998
asked Jan 26 at 19:12
jk1998jk1998
264
asked Jan 26 at 19:12
jk1998jk1998
264
asked Jan 26 at 19:12
jk1998jk1998
264
264
2
$begingroup$
This is not true.
$endgroup$
– verret
Jan 26 at 19:21
1
$begingroup$
There is no reason $cdc^{-1}$ has to be a power of $d$. If for some reason you know that $cdc^{-1} = d^k$ for some $k$, then conjugating by $c$ twice gives $d = cd^kc^{-1} = (cdc^{-1})^k = (d^k)^k = d^{k^2}$, so $1 equiv k^2 bmod 5$ and thus $k equiv pm 1 bmod 5$. More generally, if $d$ has order $m$ and for some reason you know $cdc^{-1} = d^k$ then $k^2 equiv 1 bmod m$. Generally speaking there can be more solutions to $k^2 equiv 1 bmod m$ than just $pm 1 bmod m$, although those are the only solutions when $m$ is prime. The main problem is $cdc^{-1}$ need not be a power of $d$.
$endgroup$
– KCd
Jan 26 at 19:27
$begingroup$
"What's a good way to start?". A good way to start would be to check again the source of the problem and to write an exact formulation of the problem. Reading the title I have the impression that you wrote this down out of your memory, and some things are missing.
$endgroup$
– Dietrich Burde
Jan 26 at 19:37
$begingroup$
I’m sorry, I forgot that G needs to be a group of 10 elements
$endgroup$
– jk1998
Jan 26 at 19:41
$begingroup$
Hint: What are the possible groups with $10$ elements (up to isomorphism)?
$endgroup$
– Somos
Jan 26 at 19:42
|
show 2 more comments
2
$begingroup$
This is not true.
$endgroup$
– verret
Jan 26 at 19:21
1
$begingroup$
There is no reason $cdc^{-1}$ has to be a power of $d$. If for some reason you know that $cdc^{-1} = d^k$ for some $k$, then conjugating by $c$ twice gives $d = cd^kc^{-1} = (cdc^{-1})^k = (d^k)^k = d^{k^2}$, so $1 equiv k^2 bmod 5$ and thus $k equiv pm 1 bmod 5$. More generally, if $d$ has order $m$ and for some reason you know $cdc^{-1} = d^k$ then $k^2 equiv 1 bmod m$. Generally speaking there can be more solutions to $k^2 equiv 1 bmod m$ than just $pm 1 bmod m$, although those are the only solutions when $m$ is prime. The main problem is $cdc^{-1}$ need not be a power of $d$.
$endgroup$
– KCd
Jan 26 at 19:27
$begingroup$
"What's a good way to start?". A good way to start would be to check again the source of the problem and to write an exact formulation of the problem. Reading the title I have the impression that you wrote this down out of your memory, and some things are missing.
$endgroup$
– Dietrich Burde
Jan 26 at 19:37
$begingroup$
I’m sorry, I forgot that G needs to be a group of 10 elements
$endgroup$
– jk1998
Jan 26 at 19:41
$begingroup$
Hint: What are the possible groups with $10$ elements (up to isomorphism)?
$endgroup$
– Somos
Jan 26 at 19:42
2
2
$begingroup$
This is not true.
$endgroup$
– verret
Jan 26 at 19:21
$begingroup$
This is not true.
$endgroup$
– verret
Jan 26 at 19:21
1
1
$begingroup$
There is no reason $cdc^{-1}$ has to be a power of $d$. If for some reason you know that $cdc^{-1} = d^k$ for some $k$, then conjugating by $c$ twice gives $d = cd^kc^{-1} = (cdc^{-1})^k = (d^k)^k = d^{k^2}$, so $1 equiv k^2 bmod 5$ and thus $k equiv pm 1 bmod 5$. More generally, if $d$ has order $m$ and for some reason you know $cdc^{-1} = d^k$ then $k^2 equiv 1 bmod m$. Generally speaking there can be more solutions to $k^2 equiv 1 bmod m$ than just $pm 1 bmod m$, although those are the only solutions when $m$ is prime. The main problem is $cdc^{-1}$ need not be a power of $d$.
$endgroup$
– KCd
Jan 26 at 19:27
$begingroup$
There is no reason $cdc^{-1}$ has to be a power of $d$. If for some reason you know that $cdc^{-1} = d^k$ for some $k$, then conjugating by $c$ twice gives $d = cd^kc^{-1} = (cdc^{-1})^k = (d^k)^k = d^{k^2}$, so $1 equiv k^2 bmod 5$ and thus $k equiv pm 1 bmod 5$. More generally, if $d$ has order $m$ and for some reason you know $cdc^{-1} = d^k$ then $k^2 equiv 1 bmod m$. Generally speaking there can be more solutions to $k^2 equiv 1 bmod m$ than just $pm 1 bmod m$, although those are the only solutions when $m$ is prime. The main problem is $cdc^{-1}$ need not be a power of $d$.
$endgroup$
– KCd
Jan 26 at 19:27
$begingroup$
"What's a good way to start?". A good way to start would be to check again the source of the problem and to write an exact formulation of the problem. Reading the title I have the impression that you wrote this down out of your memory, and some things are missing.
$endgroup$
– Dietrich Burde
Jan 26 at 19:37
$begingroup$
"What's a good way to start?". A good way to start would be to check again the source of the problem and to write an exact formulation of the problem. Reading the title I have the impression that you wrote this down out of your memory, and some things are missing.
$endgroup$
– Dietrich Burde
Jan 26 at 19:37
$begingroup$
I’m sorry, I forgot that G needs to be a group of 10 elements
$endgroup$
– jk1998
Jan 26 at 19:41
$begingroup$
I’m sorry, I forgot that G needs to be a group of 10 elements
$endgroup$
– jk1998
Jan 26 at 19:41
$begingroup$
Hint: What are the possible groups with $10$ elements (up to isomorphism)?
$endgroup$
– Somos
Jan 26 at 19:42
$begingroup$
Hint: What are the possible groups with $10$ elements (up to isomorphism)?
$endgroup$
– Somos
Jan 26 at 19:42
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Hint: A subgroup of order $5$ in $G$ has index $2$, hence it must be normal. So we have $cdc^{-1}in langle drangle $, i.e., $cdc^{-1}=d^k$ for some $k$. Now conclude that $k=pm 1$.
answered Jan 26 at 19:48
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
$begingroup$
Do I still need to use the assumptions above or does it then follow simply through the fact, that the subgroup is normal? I’m kind of stuck
$endgroup$
– jk1998
Jan 26 at 20:22
$begingroup$
Use that if the group isn't abelian it's $D_5$.
$endgroup$
– Chris Custer
Jan 26 at 20:42
$begingroup$
But I need to proof I’m in $D_5$ :D
$endgroup$
– jk1998
Jan 26 at 20:42
$begingroup$
True. See this: groupprops.subwiki.org/wiki/…
$endgroup$
– Chris Custer
Jan 26 at 20:46
add a comment |
$begingroup$
I must confess that this answer is actually off-point, since shows the two hypotheses (1) and (3) are mutually exclusive, not that at least one holds.
Nevertheless, for the time being at least, I'm leaving this answer as is, since it seems to have generated an engaging discussion in the comments.
Let $e_G$ denote the identity element of $G$.
The hypothesis $vert G vert = 10$ does not appear to be necessary:
If
$cdc^{-1} = d, tag 1$
then
$cd = dc; tag 2$
now if
$d^{-1} = cdc^{-1}, tag 3$
we may write
$d^{-1} = cdc^{-1} = dcc^{-1} = d Longrightarrow d^2 = e_G; tag 4$
that is, assuming (1) and (3) both hold contradicts the hypothesis
$o(d) = 5. tag 5$
answered Jan 26 at 19:50
Robert LewisRobert Lewis
48.3k23167
$endgroup$
1
$begingroup$
But your assuming that G is commutative
$endgroup$
– jk1998
Jan 26 at 20:02
1
$begingroup$
In the first equality
$endgroup$
– jk1998
Jan 26 at 20:06
2
$begingroup$
@jk1998 Two elements $c,d$ of a group commuting does not imply that $G$ is commutative.
$endgroup$
– Dietrich Burde
Jan 26 at 20:07
1
$begingroup$
I know but, what I need to proof is that either one of the equalities hold, but not at the same time
$endgroup$
– jk1998
Jan 26 at 20:09
1
$begingroup$
@DietrichBurde If c and d commute, would not the the commutivity of G be implied in this case?
$endgroup$
– jk1998
Jan 26 at 20:21
|
show 5 more comments
$begingroup$
My solution would be the following,
Since the 5 order subgroup has index 2 it must be normal in G and has the factorgroup $Z/2Z$, therefore the commutatorgroup $[G,G]$ must be a subgroup of $<d>$, if it is trivial the first equation holds and G must be abelian, if it is the whole group the second one holds true.
answered Jan 27 at 11:18
jk1998jk1998
264
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint: A subgroup of order $5$ in $G$ has index $2$, hence it must be normal. So we have $cdc^{-1}in langle drangle $, i.e., $cdc^{-1}=d^k$ for some $k$. Now conclude that $k=pm 1$.
answered Jan 26 at 19:48
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
$begingroup$
Do I still need to use the assumptions above or does it then follow simply through the fact, that the subgroup is normal? I’m kind of stuck
$endgroup$
– jk1998
Jan 26 at 20:22
$begingroup$
Use that if the group isn't abelian it's $D_5$.
$endgroup$
– Chris Custer
Jan 26 at 20:42
$begingroup$
But I need to proof I’m in $D_5$ :D
$endgroup$
– jk1998
Jan 26 at 20:42
$begingroup$
True. See this: groupprops.subwiki.org/wiki/…
$endgroup$
– Chris Custer
Jan 26 at 20:46
add a comment |
$begingroup$
Hint: A subgroup of order $5$ in $G$ has index $2$, hence it must be normal. So we have $cdc^{-1}in langle drangle $, i.e., $cdc^{-1}=d^k$ for some $k$. Now conclude that $k=pm 1$.
answered Jan 26 at 19:48
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
$begingroup$
Do I still need to use the assumptions above or does it then follow simply through the fact, that the subgroup is normal? I’m kind of stuck
$endgroup$
– jk1998
Jan 26 at 20:22
$begingroup$
Use that if the group isn't abelian it's $D_5$.
$endgroup$
– Chris Custer
Jan 26 at 20:42
$begingroup$
But I need to proof I’m in $D_5$ :D
$endgroup$
– jk1998
Jan 26 at 20:42
$begingroup$
True. See this: groupprops.subwiki.org/wiki/…
$endgroup$
– Chris Custer
Jan 26 at 20:46
add a comment |
$begingroup$
Hint: A subgroup of order $5$ in $G$ has index $2$, hence it must be normal. So we have $cdc^{-1}in langle drangle $, i.e., $cdc^{-1}=d^k$ for some $k$. Now conclude that $k=pm 1$.
answered Jan 26 at 19:48
Dietrich BurdeDietrich Burde
81.2k648106
$endgroup$
Hint: A subgroup of order $5$ in $G$ has index $2$, hence it must be normal. So we have $cdc^{-1}in langle drangle $, i.e., $cdc^{-1}=d^k$ for some $k$. Now conclude that $k=pm 1$.
answered Jan 26 at 19:48
Dietrich BurdeDietrich Burde
81.2k648106
edited Jan 26 at 19:54
answered Jan 26 at 19:48
Dietrich BurdeDietrich Burde
81.2k648106
answered Jan 26 at 19:48
Dietrich BurdeDietrich Burde
81.2k648106
answered Jan 26 at 19:48
Dietrich BurdeDietrich Burde
81.2k648106
81.2k648106
$begingroup$
Do I still need to use the assumptions above or does it then follow simply through the fact, that the subgroup is normal? I’m kind of stuck
$endgroup$
– jk1998
Jan 26 at 20:22
$begingroup$
Use that if the group isn't abelian it's $D_5$.
$endgroup$
– Chris Custer
Jan 26 at 20:42
$begingroup$
But I need to proof I’m in $D_5$ :D
$endgroup$
– jk1998
Jan 26 at 20:42
$begingroup$
True. See this: groupprops.subwiki.org/wiki/…
$endgroup$
– Chris Custer
Jan 26 at 20:46
add a comment |
$begingroup$
Do I still need to use the assumptions above or does it then follow simply through the fact, that the subgroup is normal? I’m kind of stuck
$endgroup$
– jk1998
Jan 26 at 20:22
$begingroup$
Use that if the group isn't abelian it's $D_5$.
$endgroup$
– Chris Custer
Jan 26 at 20:42
$begingroup$
But I need to proof I’m in $D_5$ :D
$endgroup$
– jk1998
Jan 26 at 20:42
$begingroup$
True. See this: groupprops.subwiki.org/wiki/…
$endgroup$
– Chris Custer
Jan 26 at 20:46
$begingroup$
Do I still need to use the assumptions above or does it then follow simply through the fact, that the subgroup is normal? I’m kind of stuck
$endgroup$
– jk1998
Jan 26 at 20:22
$begingroup$
Do I still need to use the assumptions above or does it then follow simply through the fact, that the subgroup is normal? I’m kind of stuck
$endgroup$
– jk1998
Jan 26 at 20:22
$begingroup$
Use that if the group isn't abelian it's $D_5$.
$endgroup$
– Chris Custer
Jan 26 at 20:42
$begingroup$
Use that if the group isn't abelian it's $D_5$.
$endgroup$
– Chris Custer
Jan 26 at 20:42
$begingroup$
But I need to proof I’m in $D_5$ :D
$endgroup$
– jk1998
Jan 26 at 20:42
$begingroup$
But I need to proof I’m in $D_5$ :D
$endgroup$
– jk1998
Jan 26 at 20:42
$begingroup$
True. See this: groupprops.subwiki.org/wiki/…
$endgroup$
– Chris Custer
Jan 26 at 20:46
$begingroup$
True. See this: groupprops.subwiki.org/wiki/…
$endgroup$
– Chris Custer
Jan 26 at 20:46
add a comment |
$begingroup$
I must confess that this answer is actually off-point, since shows the two hypotheses (1) and (3) are mutually exclusive, not that at least one holds.
Nevertheless, for the time being at least, I'm leaving this answer as is, since it seems to have generated an engaging discussion in the comments.
Let $e_G$ denote the identity element of $G$.
The hypothesis $vert G vert = 10$ does not appear to be necessary:
If
$cdc^{-1} = d, tag 1$
then
$cd = dc; tag 2$
now if
$d^{-1} = cdc^{-1}, tag 3$
we may write
$d^{-1} = cdc^{-1} = dcc^{-1} = d Longrightarrow d^2 = e_G; tag 4$
that is, assuming (1) and (3) both hold contradicts the hypothesis
$o(d) = 5. tag 5$
answered Jan 26 at 19:50
Robert LewisRobert Lewis
48.3k23167
$endgroup$
1
$begingroup$
But your assuming that G is commutative
$endgroup$
– jk1998
Jan 26 at 20:02
1
$begingroup$
In the first equality
$endgroup$
– jk1998
Jan 26 at 20:06
2
$begingroup$
@jk1998 Two elements $c,d$ of a group commuting does not imply that $G$ is commutative.
$endgroup$
– Dietrich Burde
Jan 26 at 20:07
1
$begingroup$
I know but, what I need to proof is that either one of the equalities hold, but not at the same time
$endgroup$
– jk1998
Jan 26 at 20:09
1
$begingroup$
@DietrichBurde If c and d commute, would not the the commutivity of G be implied in this case?
$endgroup$
– jk1998
Jan 26 at 20:21
|
show 5 more comments
$begingroup$
I must confess that this answer is actually off-point, since shows the two hypotheses (1) and (3) are mutually exclusive, not that at least one holds.
Nevertheless, for the time being at least, I'm leaving this answer as is, since it seems to have generated an engaging discussion in the comments.
Let $e_G$ denote the identity element of $G$.
The hypothesis $vert G vert = 10$ does not appear to be necessary:
If
$cdc^{-1} = d, tag 1$
then
$cd = dc; tag 2$
now if
$d^{-1} = cdc^{-1}, tag 3$
we may write
$d^{-1} = cdc^{-1} = dcc^{-1} = d Longrightarrow d^2 = e_G; tag 4$
that is, assuming (1) and (3) both hold contradicts the hypothesis
$o(d) = 5. tag 5$
answered Jan 26 at 19:50
Robert LewisRobert Lewis
48.3k23167
$endgroup$
1
$begingroup$
But your assuming that G is commutative
$endgroup$
– jk1998
Jan 26 at 20:02
1
$begingroup$
In the first equality
$endgroup$
– jk1998
Jan 26 at 20:06
2
$begingroup$
@jk1998 Two elements $c,d$ of a group commuting does not imply that $G$ is commutative.
$endgroup$
– Dietrich Burde
Jan 26 at 20:07
1
$begingroup$
I know but, what I need to proof is that either one of the equalities hold, but not at the same time
$endgroup$
– jk1998
Jan 26 at 20:09
1
$begingroup$
@DietrichBurde If c and d commute, would not the the commutivity of G be implied in this case?
$endgroup$
– jk1998
Jan 26 at 20:21
|
show 5 more comments
$begingroup$
I must confess that this answer is actually off-point, since shows the two hypotheses (1) and (3) are mutually exclusive, not that at least one holds.
Nevertheless, for the time being at least, I'm leaving this answer as is, since it seems to have generated an engaging discussion in the comments.
Let $e_G$ denote the identity element of $G$.
The hypothesis $vert G vert = 10$ does not appear to be necessary:
If
$cdc^{-1} = d, tag 1$
then
$cd = dc; tag 2$
now if
$d^{-1} = cdc^{-1}, tag 3$
we may write
$d^{-1} = cdc^{-1} = dcc^{-1} = d Longrightarrow d^2 = e_G; tag 4$
that is, assuming (1) and (3) both hold contradicts the hypothesis
$o(d) = 5. tag 5$
answered Jan 26 at 19:50
Robert LewisRobert Lewis
48.3k23167
$endgroup$
I must confess that this answer is actually off-point, since shows the two hypotheses (1) and (3) are mutually exclusive, not that at least one holds.
Nevertheless, for the time being at least, I'm leaving this answer as is, since it seems to have generated an engaging discussion in the comments.
Let $e_G$ denote the identity element of $G$.
The hypothesis $vert G vert = 10$ does not appear to be necessary:
If
$cdc^{-1} = d, tag 1$
then
$cd = dc; tag 2$
now if
$d^{-1} = cdc^{-1}, tag 3$
we may write
$d^{-1} = cdc^{-1} = dcc^{-1} = d Longrightarrow d^2 = e_G; tag 4$
that is, assuming (1) and (3) both hold contradicts the hypothesis
$o(d) = 5. tag 5$
answered Jan 26 at 19:50
Robert LewisRobert Lewis
48.3k23167
edited Jan 26 at 20:29
answered Jan 26 at 19:50
Robert LewisRobert Lewis
48.3k23167
answered Jan 26 at 19:50
Robert LewisRobert Lewis
48.3k23167
answered Jan 26 at 19:50
Robert LewisRobert Lewis
48.3k23167
48.3k23167
1
$begingroup$
But your assuming that G is commutative
$endgroup$
– jk1998
Jan 26 at 20:02
1
$begingroup$
In the first equality
$endgroup$
– jk1998
Jan 26 at 20:06
2
$begingroup$
@jk1998 Two elements $c,d$ of a group commuting does not imply that $G$ is commutative.
$endgroup$
– Dietrich Burde
Jan 26 at 20:07
1
$begingroup$
I know but, what I need to proof is that either one of the equalities hold, but not at the same time
$endgroup$
– jk1998
Jan 26 at 20:09
1
$begingroup$
@DietrichBurde If c and d commute, would not the the commutivity of G be implied in this case?
$endgroup$
– jk1998
Jan 26 at 20:21
|
show 5 more comments
1
$begingroup$
But your assuming that G is commutative
$endgroup$
– jk1998
Jan 26 at 20:02
1
$begingroup$
In the first equality
$endgroup$
– jk1998
Jan 26 at 20:06
2
$begingroup$
@jk1998 Two elements $c,d$ of a group commuting does not imply that $G$ is commutative.
$endgroup$
– Dietrich Burde
Jan 26 at 20:07
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I know but, what I need to proof is that either one of the equalities hold, but not at the same time
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– jk1998
Jan 26 at 20:09
1
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@DietrichBurde If c and d commute, would not the the commutivity of G be implied in this case?
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– jk1998
Jan 26 at 20:21
1
1
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But your assuming that G is commutative
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– jk1998
Jan 26 at 20:02
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But your assuming that G is commutative
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– jk1998
Jan 26 at 20:02
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1
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In the first equality
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– jk1998
Jan 26 at 20:06
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In the first equality
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– jk1998
Jan 26 at 20:06
2
2
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@jk1998 Two elements $c,d$ of a group commuting does not imply that $G$ is commutative.
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– Dietrich Burde
Jan 26 at 20:07
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@jk1998 Two elements $c,d$ of a group commuting does not imply that $G$ is commutative.
$endgroup$
– Dietrich Burde
Jan 26 at 20:07
1
1
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I know but, what I need to proof is that either one of the equalities hold, but not at the same time
$endgroup$
– jk1998
Jan 26 at 20:09
$begingroup$
I know but, what I need to proof is that either one of the equalities hold, but not at the same time
$endgroup$
– jk1998
Jan 26 at 20:09
1
1
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@DietrichBurde If c and d commute, would not the the commutivity of G be implied in this case?
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– jk1998
Jan 26 at 20:21
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@DietrichBurde If c and d commute, would not the the commutivity of G be implied in this case?
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– jk1998
Jan 26 at 20:21
|
show 5 more comments
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My solution would be the following,
Since the 5 order subgroup has index 2 it must be normal in G and has the factorgroup $Z/2Z$, therefore the commutatorgroup $[G,G]$ must be a subgroup of $<d>$, if it is trivial the first equation holds and G must be abelian, if it is the whole group the second one holds true.
answered Jan 27 at 11:18
jk1998jk1998
264
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add a comment |
$begingroup$
My solution would be the following,
Since the 5 order subgroup has index 2 it must be normal in G and has the factorgroup $Z/2Z$, therefore the commutatorgroup $[G,G]$ must be a subgroup of $<d>$, if it is trivial the first equation holds and G must be abelian, if it is the whole group the second one holds true.
answered Jan 27 at 11:18
jk1998jk1998
264
$endgroup$
add a comment |
$begingroup$
My solution would be the following,
Since the 5 order subgroup has index 2 it must be normal in G and has the factorgroup $Z/2Z$, therefore the commutatorgroup $[G,G]$ must be a subgroup of $<d>$, if it is trivial the first equation holds and G must be abelian, if it is the whole group the second one holds true.
answered Jan 27 at 11:18
jk1998jk1998
264
$endgroup$
My solution would be the following,
Since the 5 order subgroup has index 2 it must be normal in G and has the factorgroup $Z/2Z$, therefore the commutatorgroup $[G,G]$ must be a subgroup of $<d>$, if it is trivial the first equation holds and G must be abelian, if it is the whole group the second one holds true.
answered Jan 27 at 11:18
jk1998jk1998
264
answered Jan 27 at 11:18
jk1998jk1998
264
answered Jan 27 at 11:18
jk1998jk1998
264
answered Jan 27 at 11:18
jk1998jk1998
264
264
add a comment |
add a comment |
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This is not true.
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– verret
Jan 26 at 19:21
1
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There is no reason $cdc^{-1}$ has to be a power of $d$. If for some reason you know that $cdc^{-1} = d^k$ for some $k$, then conjugating by $c$ twice gives $d = cd^kc^{-1} = (cdc^{-1})^k = (d^k)^k = d^{k^2}$, so $1 equiv k^2 bmod 5$ and thus $k equiv pm 1 bmod 5$. More generally, if $d$ has order $m$ and for some reason you know $cdc^{-1} = d^k$ then $k^2 equiv 1 bmod m$. Generally speaking there can be more solutions to $k^2 equiv 1 bmod m$ than just $pm 1 bmod m$, although those are the only solutions when $m$ is prime. The main problem is $cdc^{-1}$ need not be a power of $d$.
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– KCd
Jan 26 at 19:27
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"What's a good way to start?". A good way to start would be to check again the source of the problem and to write an exact formulation of the problem. Reading the title I have the impression that you wrote this down out of your memory, and some things are missing.
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– Dietrich Burde
Jan 26 at 19:37
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I’m sorry, I forgot that G needs to be a group of 10 elements
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– jk1998
Jan 26 at 19:41
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Hint: What are the possible groups with $10$ elements (up to isomorphism)?
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– Somos
Jan 26 at 19:42