Mixing
If a prime $p$ divides the order of a group $G$ and $p^2 > |G|$, then there is a normal subgroup of order...
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I was wondering if there were a way to prove this without invoking the full force of the Sylow theorems. Here is my attempt:
Suppose $G$ is a group, a $p$ prime divides its order, and $p^2 > |G|$. By Cauchy's theorem, there exists a subgroup $H$ such that $|H| = p$. Suppose there exists another subgroup $K$, $|K| = p$, $K neq H$. Since $H$ and $K$ are of prime order, any nontrivial element in each generates the entire subgroup, so we must have $H cap K = {e}$. By counting argument, we have $$ |HK| = frac{|H| |K|}{|H cap K|} = p^2 > |G|,$$ contradicting $HK subseteq G$. Therefore, such a $K$ does not exist, and we conclude $H$ is the unique subgroup of $G$ of order $p$.
This uniqueness implies $H triangleleft G$: for all $g in G$, $gHg^{-1}$ is a subgroup of order $p$, so we must have $gHg^{-1} = H, forall g in G$.
Is there anything inconsistent in this proof? If not, are there even more elementary proofs?
abstract-algebra group-theory proof-verification
asked Jan 21 at 0:22
t.yt.y
427
$endgroup$
add a comment |
$begingroup$
I was wondering if there were a way to prove this without invoking the full force of the Sylow theorems. Here is my attempt:
Suppose $G$ is a group, a $p$ prime divides its order, and $p^2 > |G|$. By Cauchy's theorem, there exists a subgroup $H$ such that $|H| = p$. Suppose there exists another subgroup $K$, $|K| = p$, $K neq H$. Since $H$ and $K$ are of prime order, any nontrivial element in each generates the entire subgroup, so we must have $H cap K = {e}$. By counting argument, we have $$ |HK| = frac{|H| |K|}{|H cap K|} = p^2 > |G|,$$ contradicting $HK subseteq G$. Therefore, such a $K$ does not exist, and we conclude $H$ is the unique subgroup of $G$ of order $p$.
This uniqueness implies $H triangleleft G$: for all $g in G$, $gHg^{-1}$ is a subgroup of order $p$, so we must have $gHg^{-1} = H, forall g in G$.
Is there anything inconsistent in this proof? If not, are there even more elementary proofs?
abstract-algebra group-theory proof-verification
asked Jan 21 at 0:22
t.yt.y
427
$endgroup$
add a comment |
$begingroup$
I was wondering if there were a way to prove this without invoking the full force of the Sylow theorems. Here is my attempt:
Suppose $G$ is a group, a $p$ prime divides its order, and $p^2 > |G|$. By Cauchy's theorem, there exists a subgroup $H$ such that $|H| = p$. Suppose there exists another subgroup $K$, $|K| = p$, $K neq H$. Since $H$ and $K$ are of prime order, any nontrivial element in each generates the entire subgroup, so we must have $H cap K = {e}$. By counting argument, we have $$ |HK| = frac{|H| |K|}{|H cap K|} = p^2 > |G|,$$ contradicting $HK subseteq G$. Therefore, such a $K$ does not exist, and we conclude $H$ is the unique subgroup of $G$ of order $p$.
This uniqueness implies $H triangleleft G$: for all $g in G$, $gHg^{-1}$ is a subgroup of order $p$, so we must have $gHg^{-1} = H, forall g in G$.
Is there anything inconsistent in this proof? If not, are there even more elementary proofs?
abstract-algebra group-theory proof-verification
asked Jan 21 at 0:22
t.yt.y
427
$endgroup$
I was wondering if there were a way to prove this without invoking the full force of the Sylow theorems. Here is my attempt:
Suppose $G$ is a group, a $p$ prime divides its order, and $p^2 > |G|$. By Cauchy's theorem, there exists a subgroup $H$ such that $|H| = p$. Suppose there exists another subgroup $K$, $|K| = p$, $K neq H$. Since $H$ and $K$ are of prime order, any nontrivial element in each generates the entire subgroup, so we must have $H cap K = {e}$. By counting argument, we have $$ |HK| = frac{|H| |K|}{|H cap K|} = p^2 > |G|,$$ contradicting $HK subseteq G$. Therefore, such a $K$ does not exist, and we conclude $H$ is the unique subgroup of $G$ of order $p$.
This uniqueness implies $H triangleleft G$: for all $g in G$, $gHg^{-1}$ is a subgroup of order $p$, so we must have $gHg^{-1} = H, forall g in G$.
Is there anything inconsistent in this proof? If not, are there even more elementary proofs?
abstract-algebra group-theory proof-verification
abstract-algebra group-theory proof-verification
asked Jan 21 at 0:22
t.yt.y
427
asked Jan 21 at 0:22
t.yt.y
427
asked Jan 21 at 0:22
t.yt.y
427
asked Jan 21 at 0:22
t.yt.y
427
asked Jan 21 at 0:22
t.yt.y
427
427
add a comment |
add a comment |
1 Answer
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Yes, that argument is correct.
Another way to prove it: Let $|G|=pm$ and $H$ be an order p subgroup. Then the action of $G$ on the cosets of $H$ gives a homomorphism from $G$ to $S_m$. The action is transitive on the $m$ cosets of $H$. Also, $H$ is in the kernel since p does not divide $|S_m|$. Since $H$ has index m, it must be the entire kernel, and thus normal
answered Jan 21 at 0:33
C MonsourC Monsour
6,2541325
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1 Answer
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1 Answer
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$begingroup$
Yes, that argument is correct.
Another way to prove it: Let $|G|=pm$ and $H$ be an order p subgroup. Then the action of $G$ on the cosets of $H$ gives a homomorphism from $G$ to $S_m$. The action is transitive on the $m$ cosets of $H$. Also, $H$ is in the kernel since p does not divide $|S_m|$. Since $H$ has index m, it must be the entire kernel, and thus normal
answered Jan 21 at 0:33
C MonsourC Monsour
6,2541325
$endgroup$
add a comment |
$begingroup$
Yes, that argument is correct.
Another way to prove it: Let $|G|=pm$ and $H$ be an order p subgroup. Then the action of $G$ on the cosets of $H$ gives a homomorphism from $G$ to $S_m$. The action is transitive on the $m$ cosets of $H$. Also, $H$ is in the kernel since p does not divide $|S_m|$. Since $H$ has index m, it must be the entire kernel, and thus normal
answered Jan 21 at 0:33
C MonsourC Monsour
6,2541325
$endgroup$
add a comment |
$begingroup$
Yes, that argument is correct.
Another way to prove it: Let $|G|=pm$ and $H$ be an order p subgroup. Then the action of $G$ on the cosets of $H$ gives a homomorphism from $G$ to $S_m$. The action is transitive on the $m$ cosets of $H$. Also, $H$ is in the kernel since p does not divide $|S_m|$. Since $H$ has index m, it must be the entire kernel, and thus normal
answered Jan 21 at 0:33
C MonsourC Monsour
6,2541325
$endgroup$
Yes, that argument is correct.
Another way to prove it: Let $|G|=pm$ and $H$ be an order p subgroup. Then the action of $G$ on the cosets of $H$ gives a homomorphism from $G$ to $S_m$. The action is transitive on the $m$ cosets of $H$. Also, $H$ is in the kernel since p does not divide $|S_m|$. Since $H$ has index m, it must be the entire kernel, and thus normal
answered Jan 21 at 0:33
C MonsourC Monsour
6,2541325
edited Jan 21 at 0:48
answered Jan 21 at 0:33
C MonsourC Monsour
6,2541325
answered Jan 21 at 0:33
C MonsourC Monsour
6,2541325
answered Jan 21 at 0:33
C MonsourC Monsour
6,2541325
6,2541325
add a comment |
add a comment |
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