If $H$ is a subgroup with prime index $p$ of a finite simple group $G$, then $p$ is the maximal prime $p$...

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Let $G$ be a finite simple group. Let $H$ be a subgroup of $G$ whose index is a prime $p$. Prove that $p$ is the maximal prime dividing the order of $G$ and that $p^2 nmid |G|$.

group-theory

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edited yesterday

Tianlalu

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mathnoob

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Let $G$ be a finite simple group. Let $H$ be a subgroup of $G$ whose index is a prime $p$. Prove that $p$ is the maximal prime dividing the order of $G$ and that $p^2 nmid |G|$.

group-theory

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edited yesterday

Tianlalu

2,594632

asked yesterday

mathnoob

58311

put on hold as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Rebellos, jgon, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.

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up vote
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up vote
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down vote

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1

Let $G$ be a finite simple group. Let $H$ be a subgroup of $G$ whose index is a prime $p$. Prove that $p$ is the maximal prime dividing the order of $G$ and that $p^2 nmid |G|$.

group-theory

share|cite|improve this question

edited yesterday

Tianlalu

2,594632

asked yesterday

mathnoob

58311

Let $G$ be a finite simple group. Let $H$ be a subgroup of $G$ whose index is a prime $p$. Prove that $p$ is the maximal prime dividing the order of $G$ and that $p^2 nmid |G|$.

group-theory

group-theory

share|cite|improve this question

edited yesterday

Tianlalu

2,594632

asked yesterday

mathnoob

58311

share|cite|improve this question

edited yesterday

Tianlalu

2,594632

asked yesterday

mathnoob

58311

share|cite|improve this question

share|cite|improve this question

edited yesterday

Tianlalu

2,594632

edited yesterday

Tianlalu

2,594632

edited yesterday

Tianlalu

2,594632

2,594632

asked yesterday

mathnoob

58311

asked yesterday

mathnoob

58311

asked yesterday

mathnoob

58311

58311

put on hold as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Rebellos, jgon, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Derek Holt, amWhy, Rebellos, jgon, user10354138 yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Rebellos, jgon, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.

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1 Answer
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Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an homomorphism $f:G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.

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edited yesterday

Santana Afton

2,4642629

answered yesterday

mathnoob

58311

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an homomorphism $f:G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.

share|cite|improve this answer

edited yesterday

Santana Afton

2,4642629

answered yesterday

mathnoob

58311

add a comment | 

up vote
2
down vote

Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an homomorphism $f:G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.

share|cite|improve this answer

edited yesterday

Santana Afton

2,4642629

answered yesterday

mathnoob

58311

add a comment | 

up vote
2
down vote

up vote
2
down vote

Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an homomorphism $f:G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.

share|cite|improve this answer

edited yesterday

Santana Afton

2,4642629

answered yesterday

mathnoob

58311

Solution: Consider the transitive action of $G$ on the left cosets of $H$. This induces an homomorphism $f:G rightarrow S_{p}$. But $G$ is simple, so $ker(f)$ is trivial or all of $G$. Since $f$ is not the zero map $G$ injects into $S_p$, so $|G|| p!$. That says that any prime decomposition of $|G|$ is less than or equals to $p$. Also $p^2 nmid p!$ so $p^2 nmid $|G|$.

share|cite|improve this answer

edited yesterday

Santana Afton

2,4642629

answered yesterday

mathnoob

58311

share|cite|improve this answer

share|cite|improve this answer

edited yesterday

Santana Afton

2,4642629

edited yesterday

Santana Afton

2,4642629

edited yesterday

Santana Afton

2,4642629

2,4642629

answered yesterday

mathnoob

58311

answered yesterday

mathnoob

58311

answered yesterday

mathnoob

58311

58311

add a comment | 

add a comment | 

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