Mixing
Prove there is only one subgroup of order $p$
Suppose I have a group $G$ of order $|G|=pq, q < p$ where $p,q$ are primes. Also, suppose that $a$ has order $p$, then prove that for any $bnot in langle a rangle$, the order of $b$ is not $p$.
I know there is a way to do it with Sylow's theorems, but since we didn't cover Sylow theorems, I was hoping for another approach perhaps?
group-theory finite-groups
edited Nov 22 '18 at 10:39
the_fox
2,45711431
asked Nov 22 '18 at 7:33
AspiringMatAspiringMat
540518
|
show 5 more comments
Suppose I have a group $G$ of order $|G|=pq, q < p$ where $p,q$ are primes. Also, suppose that $a$ has order $p$, then prove that for any $bnot in langle a rangle$, the order of $b$ is not $p$.
I know there is a way to do it with Sylow's theorems, but since we didn't cover Sylow theorems, I was hoping for another approach perhaps?
group-theory finite-groups
edited Nov 22 '18 at 10:39
the_fox
2,45711431
asked Nov 22 '18 at 7:33
AspiringMatAspiringMat
540518
$bnot inlangle a rangleRightarrow o(a) nmid o(b)$. So $pnmid o(b) $
– 1ENİGMA1
Nov 22 '18 at 7:41
@1ENİGMA1 Why??
– AspiringMat
Nov 22 '18 at 7:42
I don't see why $b not in <a> Rightarrow o(a)|o(b)$ right away?
– AspiringMat
Nov 22 '18 at 7:43
The claim is false if you only demand $qle p$. Just consider $G=(Bbb Z/pBbb Z)^2$
– Hagen von Eitzen
Nov 22 '18 at 7:43
2
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, so $|langle branglelangle a rangle| ge p^2 > |G|$, contradiction.
– Derek Holt
Nov 22 '18 at 7:51
|
show 5 more comments
Suppose I have a group $G$ of order $|G|=pq, q < p$ where $p,q$ are primes. Also, suppose that $a$ has order $p$, then prove that for any $bnot in langle a rangle$, the order of $b$ is not $p$.
I know there is a way to do it with Sylow's theorems, but since we didn't cover Sylow theorems, I was hoping for another approach perhaps?
group-theory finite-groups
edited Nov 22 '18 at 10:39
the_fox
2,45711431
asked Nov 22 '18 at 7:33
AspiringMatAspiringMat
540518
Suppose I have a group $G$ of order $|G|=pq, q < p$ where $p,q$ are primes. Also, suppose that $a$ has order $p$, then prove that for any $bnot in langle a rangle$, the order of $b$ is not $p$.
I know there is a way to do it with Sylow's theorems, but since we didn't cover Sylow theorems, I was hoping for another approach perhaps?
group-theory finite-groups
group-theory finite-groups
edited Nov 22 '18 at 10:39
the_fox
2,45711431
asked Nov 22 '18 at 7:33
AspiringMatAspiringMat
540518
edited Nov 22 '18 at 10:39
the_fox
2,45711431
asked Nov 22 '18 at 7:33
AspiringMatAspiringMat
540518
edited Nov 22 '18 at 10:39
the_fox
2,45711431
edited Nov 22 '18 at 10:39
the_fox
2,45711431
edited Nov 22 '18 at 10:39
the_fox
2,45711431
2,45711431
asked Nov 22 '18 at 7:33
AspiringMatAspiringMat
540518
asked Nov 22 '18 at 7:33
AspiringMatAspiringMat
540518
asked Nov 22 '18 at 7:33
AspiringMatAspiringMat
540518
540518
$bnot inlangle a rangleRightarrow o(a) nmid o(b)$. So $pnmid o(b) $
– 1ENİGMA1
Nov 22 '18 at 7:41
@1ENİGMA1 Why??
– AspiringMat
Nov 22 '18 at 7:42
I don't see why $b not in <a> Rightarrow o(a)|o(b)$ right away?
– AspiringMat
Nov 22 '18 at 7:43
The claim is false if you only demand $qle p$. Just consider $G=(Bbb Z/pBbb Z)^2$
– Hagen von Eitzen
Nov 22 '18 at 7:43
2
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, so $|langle branglelangle a rangle| ge p^2 > |G|$, contradiction.
– Derek Holt
Nov 22 '18 at 7:51
|
show 5 more comments
$bnot inlangle a rangleRightarrow o(a) nmid o(b)$. So $pnmid o(b) $
– 1ENİGMA1
Nov 22 '18 at 7:41
@1ENİGMA1 Why??
– AspiringMat
Nov 22 '18 at 7:42
I don't see why $b not in <a> Rightarrow o(a)|o(b)$ right away?
– AspiringMat
Nov 22 '18 at 7:43
The claim is false if you only demand $qle p$. Just consider $G=(Bbb Z/pBbb Z)^2$
– Hagen von Eitzen
Nov 22 '18 at 7:43
2
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, so $|langle branglelangle a rangle| ge p^2 > |G|$, contradiction.
– Derek Holt
Nov 22 '18 at 7:51
$bnot inlangle a rangleRightarrow o(a) nmid o(b)$. So $pnmid o(b) $
– 1ENİGMA1
Nov 22 '18 at 7:41
$bnot inlangle a rangleRightarrow o(a) nmid o(b)$. So $pnmid o(b) $
– 1ENİGMA1
Nov 22 '18 at 7:41
@1ENİGMA1 Why??
– AspiringMat
Nov 22 '18 at 7:42
@1ENİGMA1 Why??
– AspiringMat
Nov 22 '18 at 7:42
I don't see why $b not in <a> Rightarrow o(a)|o(b)$ right away?
– AspiringMat
Nov 22 '18 at 7:43
I don't see why $b not in <a> Rightarrow o(a)|o(b)$ right away?
– AspiringMat
Nov 22 '18 at 7:43
The claim is false if you only demand $qle p$. Just consider $G=(Bbb Z/pBbb Z)^2$
– Hagen von Eitzen
Nov 22 '18 at 7:43
The claim is false if you only demand $qle p$. Just consider $G=(Bbb Z/pBbb Z)^2$
– Hagen von Eitzen
Nov 22 '18 at 7:43
2
2
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, so $|langle branglelangle a rangle| ge p^2 > |G|$, contradiction.
– Derek Holt
Nov 22 '18 at 7:51
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, so $|langle branglelangle a rangle| ge p^2 > |G|$, contradiction.
– Derek Holt
Nov 22 '18 at 7:51
|
show 5 more comments
2 Answers
2
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I'll make my comment into an answer.
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, and so $|langle b ranglelangle a rangle|=p^2 > |G|$, contradiction.
Since you are looking for an elementary proof, I will justify the claim that $|langle b ranglelangle a rangle|=p^2$. If not, then there exist $i,j,k,l in [0 .. p-1]$ with $(i,j) ne (k,l)$ and $a^ib^j = a^kb^l$, but then $a^{i-k} = b^{l-j}$ is a nontrivial element of $langle b rangle cap langle a rangle$.
There is a general result that, for finite subgroups $A$ and $B$ of a group $G$, we have $|AB| = |A||B|/|A cap B|$
answered Nov 22 '18 at 8:03
Derek HoltDerek Holt
52.7k53570
add a comment |
More generally, if $q$ is the smallest prime dividing $|G|$, then any subgroup $H$ of index $q$ is normal.$^1$
Now in the given problem, what can you say about the image of $b$ in $G/langle arangle$?
$^1$ Namely, $G$ acts by left multiplication on the set $X={,gHmid gin G,}$ of the $q$ cosets, which gives us a homomorphism $phicolon Gto S_q$. Clearly, $kerphisubseteq H$.
Note that $H$ leaves the point $Hin X$ fixed, hence is in fact mapped by $phi$ into $S_{q-1}$. As $|S_{q-1}|$ is co-prime to $|G|$, we must have $Hsubseteq kerphi$.
answered Nov 22 '18 at 7:46
Hagen von EitzenHagen von Eitzen
276k21269496
3
But surely you need to justify your claim that the subgroup of index $q$ is normal.
– Derek Holt
Nov 22 '18 at 7:49
@DerekHolt This ought to have been shown shortly after the introduction of normal subgroups, typically as an afterthought to the beginner's exercise about the case of index $2$. With the Sylow theorems not covered but having been heard of, my guess was this was known ...
– Hagen von Eitzen
Nov 22 '18 at 8:39
add a comment |
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2 Answers
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2 Answers
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I'll make my comment into an answer.
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, and so $|langle b ranglelangle a rangle|=p^2 > |G|$, contradiction.
Since you are looking for an elementary proof, I will justify the claim that $|langle b ranglelangle a rangle|=p^2$. If not, then there exist $i,j,k,l in [0 .. p-1]$ with $(i,j) ne (k,l)$ and $a^ib^j = a^kb^l$, but then $a^{i-k} = b^{l-j}$ is a nontrivial element of $langle b rangle cap langle a rangle$.
There is a general result that, for finite subgroups $A$ and $B$ of a group $G$, we have $|AB| = |A||B|/|A cap B|$
answered Nov 22 '18 at 8:03
Derek HoltDerek Holt
52.7k53570
add a comment |
I'll make my comment into an answer.
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, and so $|langle b ranglelangle a rangle|=p^2 > |G|$, contradiction.
Since you are looking for an elementary proof, I will justify the claim that $|langle b ranglelangle a rangle|=p^2$. If not, then there exist $i,j,k,l in [0 .. p-1]$ with $(i,j) ne (k,l)$ and $a^ib^j = a^kb^l$, but then $a^{i-k} = b^{l-j}$ is a nontrivial element of $langle b rangle cap langle a rangle$.
There is a general result that, for finite subgroups $A$ and $B$ of a group $G$, we have $|AB| = |A||B|/|A cap B|$
answered Nov 22 '18 at 8:03
Derek HoltDerek Holt
52.7k53570
add a comment |
I'll make my comment into an answer.
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, and so $|langle b ranglelangle a rangle|=p^2 > |G|$, contradiction.
Since you are looking for an elementary proof, I will justify the claim that $|langle b ranglelangle a rangle|=p^2$. If not, then there exist $i,j,k,l in [0 .. p-1]$ with $(i,j) ne (k,l)$ and $a^ib^j = a^kb^l$, but then $a^{i-k} = b^{l-j}$ is a nontrivial element of $langle b rangle cap langle a rangle$.
There is a general result that, for finite subgroups $A$ and $B$ of a group $G$, we have $|AB| = |A||B|/|A cap B|$
answered Nov 22 '18 at 8:03
Derek HoltDerek Holt
52.7k53570
I'll make my comment into an answer.
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, and so $|langle b ranglelangle a rangle|=p^2 > |G|$, contradiction.
Since you are looking for an elementary proof, I will justify the claim that $|langle b ranglelangle a rangle|=p^2$. If not, then there exist $i,j,k,l in [0 .. p-1]$ with $(i,j) ne (k,l)$ and $a^ib^j = a^kb^l$, but then $a^{i-k} = b^{l-j}$ is a nontrivial element of $langle b rangle cap langle a rangle$.
There is a general result that, for finite subgroups $A$ and $B$ of a group $G$, we have $|AB| = |A||B|/|A cap B|$
answered Nov 22 '18 at 8:03
Derek HoltDerek Holt
52.7k53570
answered Nov 22 '18 at 8:03
Derek HoltDerek Holt
52.7k53570
answered Nov 22 '18 at 8:03
Derek HoltDerek Holt
52.7k53570
answered Nov 22 '18 at 8:03
Derek HoltDerek Holt
52.7k53570
52.7k53570
add a comment |
add a comment |
More generally, if $q$ is the smallest prime dividing $|G|$, then any subgroup $H$ of index $q$ is normal.$^1$
Now in the given problem, what can you say about the image of $b$ in $G/langle arangle$?
$^1$ Namely, $G$ acts by left multiplication on the set $X={,gHmid gin G,}$ of the $q$ cosets, which gives us a homomorphism $phicolon Gto S_q$. Clearly, $kerphisubseteq H$.
Note that $H$ leaves the point $Hin X$ fixed, hence is in fact mapped by $phi$ into $S_{q-1}$. As $|S_{q-1}|$ is co-prime to $|G|$, we must have $Hsubseteq kerphi$.
answered Nov 22 '18 at 7:46
Hagen von EitzenHagen von Eitzen
276k21269496
3
But surely you need to justify your claim that the subgroup of index $q$ is normal.
– Derek Holt
Nov 22 '18 at 7:49
@DerekHolt This ought to have been shown shortly after the introduction of normal subgroups, typically as an afterthought to the beginner's exercise about the case of index $2$. With the Sylow theorems not covered but having been heard of, my guess was this was known ...
– Hagen von Eitzen
Nov 22 '18 at 8:39
add a comment |
More generally, if $q$ is the smallest prime dividing $|G|$, then any subgroup $H$ of index $q$ is normal.$^1$
Now in the given problem, what can you say about the image of $b$ in $G/langle arangle$?
$^1$ Namely, $G$ acts by left multiplication on the set $X={,gHmid gin G,}$ of the $q$ cosets, which gives us a homomorphism $phicolon Gto S_q$. Clearly, $kerphisubseteq H$.
Note that $H$ leaves the point $Hin X$ fixed, hence is in fact mapped by $phi$ into $S_{q-1}$. As $|S_{q-1}|$ is co-prime to $|G|$, we must have $Hsubseteq kerphi$.
answered Nov 22 '18 at 7:46
Hagen von EitzenHagen von Eitzen
276k21269496
3
But surely you need to justify your claim that the subgroup of index $q$ is normal.
– Derek Holt
Nov 22 '18 at 7:49
@DerekHolt This ought to have been shown shortly after the introduction of normal subgroups, typically as an afterthought to the beginner's exercise about the case of index $2$. With the Sylow theorems not covered but having been heard of, my guess was this was known ...
– Hagen von Eitzen
Nov 22 '18 at 8:39
add a comment |
More generally, if $q$ is the smallest prime dividing $|G|$, then any subgroup $H$ of index $q$ is normal.$^1$
Now in the given problem, what can you say about the image of $b$ in $G/langle arangle$?
$^1$ Namely, $G$ acts by left multiplication on the set $X={,gHmid gin G,}$ of the $q$ cosets, which gives us a homomorphism $phicolon Gto S_q$. Clearly, $kerphisubseteq H$.
Note that $H$ leaves the point $Hin X$ fixed, hence is in fact mapped by $phi$ into $S_{q-1}$. As $|S_{q-1}|$ is co-prime to $|G|$, we must have $Hsubseteq kerphi$.
answered Nov 22 '18 at 7:46
Hagen von EitzenHagen von Eitzen
276k21269496
More generally, if $q$ is the smallest prime dividing $|G|$, then any subgroup $H$ of index $q$ is normal.$^1$
Now in the given problem, what can you say about the image of $b$ in $G/langle arangle$?
$^1$ Namely, $G$ acts by left multiplication on the set $X={,gHmid gin G,}$ of the $q$ cosets, which gives us a homomorphism $phicolon Gto S_q$. Clearly, $kerphisubseteq H$.
Note that $H$ leaves the point $Hin X$ fixed, hence is in fact mapped by $phi$ into $S_{q-1}$. As $|S_{q-1}|$ is co-prime to $|G|$, we must have $Hsubseteq kerphi$.
answered Nov 22 '18 at 7:46
Hagen von EitzenHagen von Eitzen
276k21269496
edited Nov 22 '18 at 8:53
answered Nov 22 '18 at 7:46
Hagen von EitzenHagen von Eitzen
276k21269496
answered Nov 22 '18 at 7:46
Hagen von EitzenHagen von Eitzen
276k21269496
answered Nov 22 '18 at 7:46
Hagen von EitzenHagen von Eitzen
276k21269496
276k21269496
3
But surely you need to justify your claim that the subgroup of index $q$ is normal.
– Derek Holt
Nov 22 '18 at 7:49
@DerekHolt This ought to have been shown shortly after the introduction of normal subgroups, typically as an afterthought to the beginner's exercise about the case of index $2$. With the Sylow theorems not covered but having been heard of, my guess was this was known ...
– Hagen von Eitzen
Nov 22 '18 at 8:39
add a comment |
3
But surely you need to justify your claim that the subgroup of index $q$ is normal.
– Derek Holt
Nov 22 '18 at 7:49
@DerekHolt This ought to have been shown shortly after the introduction of normal subgroups, typically as an afterthought to the beginner's exercise about the case of index $2$. With the Sylow theorems not covered but having been heard of, my guess was this was known ...
– Hagen von Eitzen
Nov 22 '18 at 8:39
3
3
But surely you need to justify your claim that the subgroup of index $q$ is normal.
– Derek Holt
Nov 22 '18 at 7:49
But surely you need to justify your claim that the subgroup of index $q$ is normal.
– Derek Holt
Nov 22 '18 at 7:49
@DerekHolt This ought to have been shown shortly after the introduction of normal subgroups, typically as an afterthought to the beginner's exercise about the case of index $2$. With the Sylow theorems not covered but having been heard of, my guess was this was known ...
– Hagen von Eitzen
Nov 22 '18 at 8:39
@DerekHolt This ought to have been shown shortly after the introduction of normal subgroups, typically as an afterthought to the beginner's exercise about the case of index $2$. With the Sylow theorems not covered but having been heard of, my guess was this was known ...
– Hagen von Eitzen
Nov 22 '18 at 8:39
add a comment |
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$bnot inlangle a rangleRightarrow o(a) nmid o(b)$. So $pnmid o(b) $
– 1ENİGMA1
Nov 22 '18 at 7:41
@1ENİGMA1 Why??
– AspiringMat
Nov 22 '18 at 7:42
I don't see why $b not in <a> Rightarrow o(a)|o(b)$ right away?
– AspiringMat
Nov 22 '18 at 7:43
The claim is false if you only demand $qle p$. Just consider $G=(Bbb Z/pBbb Z)^2$
– Hagen von Eitzen
Nov 22 '18 at 7:43
2
If $b notin langle a rangle$ and $o(b)=p$, then $langle b rangle cap langle a rangle = {1}$, so $|langle branglelangle a rangle| ge p^2 > |G|$, contradiction.
– Derek Holt
Nov 22 '18 at 7:51