Mixing
Let $G$ be a group, and the order of the group $G$, $|G|=r$. For any $x in G$, prove that $x^{-r}=e$ (where...
Let $G$ be a finite group, and the order of the group $G$, $|G|=r$.
For any $x in G$, prove that $x^{-r}=e$ (where $e$ is a neutral in $G$).
Basically, I haven't tried anything except trying to manipulate the expression $aa^{-1}=e$ into what I'm trying to prove because I really have no idea what to do.
I would appreciate any hint, advice or etc. because I'm new in group theory and the concepts are still hard to understand.
Also, the next part of this task is this:
$G$ is a group such that $|G|=195$. Is there an element $g neq e$ such that $g^{77}=e$?
Thanks in advance!
abstract-algebra group-theory finite-groups
asked Nov 20 '18 at 22:19
mathbbandstuff
256111
add a comment |
Let $G$ be a finite group, and the order of the group $G$, $|G|=r$.
For any $x in G$, prove that $x^{-r}=e$ (where $e$ is a neutral in $G$).
Basically, I haven't tried anything except trying to manipulate the expression $aa^{-1}=e$ into what I'm trying to prove because I really have no idea what to do.
I would appreciate any hint, advice or etc. because I'm new in group theory and the concepts are still hard to understand.
Also, the next part of this task is this:
$G$ is a group such that $|G|=195$. Is there an element $g neq e$ such that $g^{77}=e$?
Thanks in advance!
abstract-algebra group-theory finite-groups
asked Nov 20 '18 at 22:19
mathbbandstuff
256111
1
For your first question, it might help to note that $(x^{-r}) = (x^{-1})^r = (x^r)^{-1}$. Either of those latter equalities, with the definition of order of a group and the fact that each $x in G$ has an inverse, should give you a push forward. I'm not 100% sure how to approach your second question though (don't have the best foundation in this topic), so I'll refrain from commenting on that.
– Eevee Trainer
Nov 20 '18 at 22:27
1
You need Lagrange's theorem.
– the_fox
Nov 20 '18 at 22:51
I'm not quite sure how to use that @the_fox
– mathbbandstuff
Nov 20 '18 at 23:00
1
For the second question, it suffices to observe that none of the factors of 77 divide the order of the group, which is necessary per Lagrange's theorem for a nonzero element to equal 1 when raised to the 77th power.
– Monstrous Moonshiner
Nov 21 '18 at 0:58
add a comment |
Let $G$ be a finite group, and the order of the group $G$, $|G|=r$.
For any $x in G$, prove that $x^{-r}=e$ (where $e$ is a neutral in $G$).
Basically, I haven't tried anything except trying to manipulate the expression $aa^{-1}=e$ into what I'm trying to prove because I really have no idea what to do.
I would appreciate any hint, advice or etc. because I'm new in group theory and the concepts are still hard to understand.
Also, the next part of this task is this:
$G$ is a group such that $|G|=195$. Is there an element $g neq e$ such that $g^{77}=e$?
Thanks in advance!
abstract-algebra group-theory finite-groups
asked Nov 20 '18 at 22:19
mathbbandstuff
256111
Let $G$ be a finite group, and the order of the group $G$, $|G|=r$.
For any $x in G$, prove that $x^{-r}=e$ (where $e$ is a neutral in $G$).
Basically, I haven't tried anything except trying to manipulate the expression $aa^{-1}=e$ into what I'm trying to prove because I really have no idea what to do.
I would appreciate any hint, advice or etc. because I'm new in group theory and the concepts are still hard to understand.
Also, the next part of this task is this:
$G$ is a group such that $|G|=195$. Is there an element $g neq e$ such that $g^{77}=e$?
Thanks in advance!
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked Nov 20 '18 at 22:19
mathbbandstuff
256111
asked Nov 20 '18 at 22:19
mathbbandstuff
256111
edited Nov 20 '18 at 22:24
asked Nov 20 '18 at 22:19
mathbbandstuff
256111
asked Nov 20 '18 at 22:19
mathbbandstuff
256111
asked Nov 20 '18 at 22:19
mathbbandstuff
256111
256111
1
For your first question, it might help to note that $(x^{-r}) = (x^{-1})^r = (x^r)^{-1}$. Either of those latter equalities, with the definition of order of a group and the fact that each $x in G$ has an inverse, should give you a push forward. I'm not 100% sure how to approach your second question though (don't have the best foundation in this topic), so I'll refrain from commenting on that.
– Eevee Trainer
Nov 20 '18 at 22:27
1
You need Lagrange's theorem.
– the_fox
Nov 20 '18 at 22:51
I'm not quite sure how to use that @the_fox
– mathbbandstuff
Nov 20 '18 at 23:00
1
For the second question, it suffices to observe that none of the factors of 77 divide the order of the group, which is necessary per Lagrange's theorem for a nonzero element to equal 1 when raised to the 77th power.
– Monstrous Moonshiner
Nov 21 '18 at 0:58
add a comment |
1
For your first question, it might help to note that $(x^{-r}) = (x^{-1})^r = (x^r)^{-1}$. Either of those latter equalities, with the definition of order of a group and the fact that each $x in G$ has an inverse, should give you a push forward. I'm not 100% sure how to approach your second question though (don't have the best foundation in this topic), so I'll refrain from commenting on that.
– Eevee Trainer
Nov 20 '18 at 22:27
1
You need Lagrange's theorem.
– the_fox
Nov 20 '18 at 22:51
I'm not quite sure how to use that @the_fox
– mathbbandstuff
Nov 20 '18 at 23:00
1
For the second question, it suffices to observe that none of the factors of 77 divide the order of the group, which is necessary per Lagrange's theorem for a nonzero element to equal 1 when raised to the 77th power.
– Monstrous Moonshiner
Nov 21 '18 at 0:58
1
1
For your first question, it might help to note that $(x^{-r}) = (x^{-1})^r = (x^r)^{-1}$. Either of those latter equalities, with the definition of order of a group and the fact that each $x in G$ has an inverse, should give you a push forward. I'm not 100% sure how to approach your second question though (don't have the best foundation in this topic), so I'll refrain from commenting on that.
– Eevee Trainer
Nov 20 '18 at 22:27
For your first question, it might help to note that $(x^{-r}) = (x^{-1})^r = (x^r)^{-1}$. Either of those latter equalities, with the definition of order of a group and the fact that each $x in G$ has an inverse, should give you a push forward. I'm not 100% sure how to approach your second question though (don't have the best foundation in this topic), so I'll refrain from commenting on that.
– Eevee Trainer
Nov 20 '18 at 22:27
1
1
You need Lagrange's theorem.
– the_fox
Nov 20 '18 at 22:51
You need Lagrange's theorem.
– the_fox
Nov 20 '18 at 22:51
I'm not quite sure how to use that @the_fox
– mathbbandstuff
Nov 20 '18 at 23:00
I'm not quite sure how to use that @the_fox
– mathbbandstuff
Nov 20 '18 at 23:00
1
1
For the second question, it suffices to observe that none of the factors of 77 divide the order of the group, which is necessary per Lagrange's theorem for a nonzero element to equal 1 when raised to the 77th power.
– Monstrous Moonshiner
Nov 21 '18 at 0:58
For the second question, it suffices to observe that none of the factors of 77 divide the order of the group, which is necessary per Lagrange's theorem for a nonzero element to equal 1 when raised to the 77th power.
– Monstrous Moonshiner
Nov 21 '18 at 0:58
add a comment |
2 Answers
2
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Suppose $o(x)=d$. This is the same as saying that $o(langle x rangle)=d$. By Lagrange's Theorem (which I assume you know), $d$ divides $r$, so $r/d$ is an integer, call it $m$. Then $x^r = (x^d)^m=e^m=e$.
For the second question: if there is such a $g$, then since $g^{77}=e$, the order of $g$ must be a divisor of $77$ (do you see why?). But, again by Lagrange, the order of $g$ must also divide $195 = |G|$, thus it must also divide $gcd(195,77)=1$. The only element of order $1$ in any group is $e$, which is a contradiction to the assumption that $g neq e$.
answered Nov 20 '18 at 23:09
the_fox
2,44411431
add a comment |
If $g^{77}=1$, then $(g^{77})^2=g^{154}=1$ and $(g^{77})^3=g^{231}=g^{231(mod 195)}=g^{36}=1$.
Notice $195=3*5*13$ and $77=7*11$ so $lcm(77,195)=77*195$ which means $77*a neq 0(mod195)$ for $a<195$. So by applying the above argument, $(g^{77})^{38}=g=1$. So contradiction.
answered Nov 20 '18 at 22:57
mathnoob
1,799422
Why $(g^{77})^{38}=g$? Sorry if I don't see something basic. Also, do you know how to solve the other problem?
– mathbbandstuff
Nov 20 '18 at 23:04
1
Sorry, I used online calculator, because 77 and 38 are inverse of each other in (mod 195). So 77*38=1(mod 195).
– mathnoob
Nov 20 '18 at 23:19
Oh, didn't think of that :) Thanks!
– mathbbandstuff
Nov 20 '18 at 23:20
add a comment |
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2 Answers
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2 Answers
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Suppose $o(x)=d$. This is the same as saying that $o(langle x rangle)=d$. By Lagrange's Theorem (which I assume you know), $d$ divides $r$, so $r/d$ is an integer, call it $m$. Then $x^r = (x^d)^m=e^m=e$.
For the second question: if there is such a $g$, then since $g^{77}=e$, the order of $g$ must be a divisor of $77$ (do you see why?). But, again by Lagrange, the order of $g$ must also divide $195 = |G|$, thus it must also divide $gcd(195,77)=1$. The only element of order $1$ in any group is $e$, which is a contradiction to the assumption that $g neq e$.
answered Nov 20 '18 at 23:09
the_fox
2,44411431
add a comment |
Suppose $o(x)=d$. This is the same as saying that $o(langle x rangle)=d$. By Lagrange's Theorem (which I assume you know), $d$ divides $r$, so $r/d$ is an integer, call it $m$. Then $x^r = (x^d)^m=e^m=e$.
For the second question: if there is such a $g$, then since $g^{77}=e$, the order of $g$ must be a divisor of $77$ (do you see why?). But, again by Lagrange, the order of $g$ must also divide $195 = |G|$, thus it must also divide $gcd(195,77)=1$. The only element of order $1$ in any group is $e$, which is a contradiction to the assumption that $g neq e$.
answered Nov 20 '18 at 23:09
the_fox
2,44411431
add a comment |
Suppose $o(x)=d$. This is the same as saying that $o(langle x rangle)=d$. By Lagrange's Theorem (which I assume you know), $d$ divides $r$, so $r/d$ is an integer, call it $m$. Then $x^r = (x^d)^m=e^m=e$.
For the second question: if there is such a $g$, then since $g^{77}=e$, the order of $g$ must be a divisor of $77$ (do you see why?). But, again by Lagrange, the order of $g$ must also divide $195 = |G|$, thus it must also divide $gcd(195,77)=1$. The only element of order $1$ in any group is $e$, which is a contradiction to the assumption that $g neq e$.
answered Nov 20 '18 at 23:09
the_fox
2,44411431
Suppose $o(x)=d$. This is the same as saying that $o(langle x rangle)=d$. By Lagrange's Theorem (which I assume you know), $d$ divides $r$, so $r/d$ is an integer, call it $m$. Then $x^r = (x^d)^m=e^m=e$.
For the second question: if there is such a $g$, then since $g^{77}=e$, the order of $g$ must be a divisor of $77$ (do you see why?). But, again by Lagrange, the order of $g$ must also divide $195 = |G|$, thus it must also divide $gcd(195,77)=1$. The only element of order $1$ in any group is $e$, which is a contradiction to the assumption that $g neq e$.
answered Nov 20 '18 at 23:09
the_fox
2,44411431
answered Nov 20 '18 at 23:09
the_fox
2,44411431
answered Nov 20 '18 at 23:09
the_fox
2,44411431
answered Nov 20 '18 at 23:09
the_fox
2,44411431
2,44411431
add a comment |
add a comment |
If $g^{77}=1$, then $(g^{77})^2=g^{154}=1$ and $(g^{77})^3=g^{231}=g^{231(mod 195)}=g^{36}=1$.
Notice $195=3*5*13$ and $77=7*11$ so $lcm(77,195)=77*195$ which means $77*a neq 0(mod195)$ for $a<195$. So by applying the above argument, $(g^{77})^{38}=g=1$. So contradiction.
answered Nov 20 '18 at 22:57
mathnoob
1,799422
Why $(g^{77})^{38}=g$? Sorry if I don't see something basic. Also, do you know how to solve the other problem?
– mathbbandstuff
Nov 20 '18 at 23:04
1
Sorry, I used online calculator, because 77 and 38 are inverse of each other in (mod 195). So 77*38=1(mod 195).
– mathnoob
Nov 20 '18 at 23:19
Oh, didn't think of that :) Thanks!
– mathbbandstuff
Nov 20 '18 at 23:20
add a comment |
If $g^{77}=1$, then $(g^{77})^2=g^{154}=1$ and $(g^{77})^3=g^{231}=g^{231(mod 195)}=g^{36}=1$.
Notice $195=3*5*13$ and $77=7*11$ so $lcm(77,195)=77*195$ which means $77*a neq 0(mod195)$ for $a<195$. So by applying the above argument, $(g^{77})^{38}=g=1$. So contradiction.
answered Nov 20 '18 at 22:57
mathnoob
1,799422
Why $(g^{77})^{38}=g$? Sorry if I don't see something basic. Also, do you know how to solve the other problem?
– mathbbandstuff
Nov 20 '18 at 23:04
1
Sorry, I used online calculator, because 77 and 38 are inverse of each other in (mod 195). So 77*38=1(mod 195).
– mathnoob
Nov 20 '18 at 23:19
Oh, didn't think of that :) Thanks!
– mathbbandstuff
Nov 20 '18 at 23:20
add a comment |
If $g^{77}=1$, then $(g^{77})^2=g^{154}=1$ and $(g^{77})^3=g^{231}=g^{231(mod 195)}=g^{36}=1$.
Notice $195=3*5*13$ and $77=7*11$ so $lcm(77,195)=77*195$ which means $77*a neq 0(mod195)$ for $a<195$. So by applying the above argument, $(g^{77})^{38}=g=1$. So contradiction.
answered Nov 20 '18 at 22:57
mathnoob
1,799422
If $g^{77}=1$, then $(g^{77})^2=g^{154}=1$ and $(g^{77})^3=g^{231}=g^{231(mod 195)}=g^{36}=1$.
Notice $195=3*5*13$ and $77=7*11$ so $lcm(77,195)=77*195$ which means $77*a neq 0(mod195)$ for $a<195$. So by applying the above argument, $(g^{77})^{38}=g=1$. So contradiction.
answered Nov 20 '18 at 22:57
mathnoob
1,799422
answered Nov 20 '18 at 22:57
mathnoob
1,799422
answered Nov 20 '18 at 22:57
mathnoob
1,799422
answered Nov 20 '18 at 22:57
mathnoob
1,799422
1,799422
Why $(g^{77})^{38}=g$? Sorry if I don't see something basic. Also, do you know how to solve the other problem?
– mathbbandstuff
Nov 20 '18 at 23:04
1
Sorry, I used online calculator, because 77 and 38 are inverse of each other in (mod 195). So 77*38=1(mod 195).
– mathnoob
Nov 20 '18 at 23:19
Oh, didn't think of that :) Thanks!
– mathbbandstuff
Nov 20 '18 at 23:20
add a comment |
Why $(g^{77})^{38}=g$? Sorry if I don't see something basic. Also, do you know how to solve the other problem?
– mathbbandstuff
Nov 20 '18 at 23:04
1
Sorry, I used online calculator, because 77 and 38 are inverse of each other in (mod 195). So 77*38=1(mod 195).
– mathnoob
Nov 20 '18 at 23:19
Oh, didn't think of that :) Thanks!
– mathbbandstuff
Nov 20 '18 at 23:20
Why $(g^{77})^{38}=g$? Sorry if I don't see something basic. Also, do you know how to solve the other problem?
– mathbbandstuff
Nov 20 '18 at 23:04
Why $(g^{77})^{38}=g$? Sorry if I don't see something basic. Also, do you know how to solve the other problem?
– mathbbandstuff
Nov 20 '18 at 23:04
1
1
Sorry, I used online calculator, because 77 and 38 are inverse of each other in (mod 195). So 77*38=1(mod 195).
– mathnoob
Nov 20 '18 at 23:19
Sorry, I used online calculator, because 77 and 38 are inverse of each other in (mod 195). So 77*38=1(mod 195).
– mathnoob
Nov 20 '18 at 23:19
Oh, didn't think of that :) Thanks!
– mathbbandstuff
Nov 20 '18 at 23:20
Oh, didn't think of that :) Thanks!
– mathbbandstuff
Nov 20 '18 at 23:20
add a comment |
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1
For your first question, it might help to note that $(x^{-r}) = (x^{-1})^r = (x^r)^{-1}$. Either of those latter equalities, with the definition of order of a group and the fact that each $x in G$ has an inverse, should give you a push forward. I'm not 100% sure how to approach your second question though (don't have the best foundation in this topic), so I'll refrain from commenting on that.
– Eevee Trainer
Nov 20 '18 at 22:27
1
You need Lagrange's theorem.
– the_fox
Nov 20 '18 at 22:51
I'm not quite sure how to use that @the_fox
– mathbbandstuff
Nov 20 '18 at 23:00
1
For the second question, it suffices to observe that none of the factors of 77 divide the order of the group, which is necessary per Lagrange's theorem for a nonzero element to equal 1 when raised to the 77th power.
– Monstrous Moonshiner
Nov 21 '18 at 0:58