Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that

if $H$ is subgroup of $G$, $H$ is a maximal subgroup of $G$ iff $[G:H]$ is prime.

I was reading a proof and the proof used this theorem to prove the statement. It's the first time that I meet this theorem !

Can any one give me its proof ?

share|improve this question
    
For the case $G$ is a finite group and violating the claim, see this complete source. Index of a maximal subgroup in a finite group. –  B. S. Feb 2 '13 at 8:53

3 Answers 3

up vote 4 down vote accepted

As pointed out by the other answers only one direction is true: If $H$ is not maximal then there is $K$ such that

$$H \subsetneqq K \subsetneqq G.$$

Then we get that $[G: H] = [G:K][K:H]$. Then condition of strict inclusion guarantees that each of these factors is not $1$. Thus $[G:H]$ is not prime.

share|improve this answer

This is true if you assume that $H$ is a normal subgroup.

If $[G:H]$ is prime, then you can prove that $H$ must be maximal by proceeding as in the other answer. If you know that $H$ is a normal maximal subgroup, then $[G:H]$ must be prime. This follows from the correspondence theorem: you can show that $G/H$ has no nontrivial proper subgroups and thus must be cyclic of prime order.

However, in general this statement is false. For example, in the alternating group $A_4$ subgroups of order $3$ are maximal and have index $4$, which is not prime.

For solvable groups we have the following result.

If $G$ is a finite solvable group and $M < G$ is a maximal subgroup, then $[G:M]$ is a power of a prime.

This is not true for groups that are not solvable, as the alternating group $A_5$ shows.

share|improve this answer

This is false for an arbitrary group $G$. If every maximal subgroup of a group $G$ has prime index, it follows that $G$ is solvable by a theorem of Hall. For example, $A_5$ has a maximal subgroup of index $6$.

share|improve this answer
    
what is the true ? –  Maths Lover Feb 2 '13 at 8:58
1  
@MathsLover Pardon me, what are you asking? –  Alexander Gruber Feb 2 '13 at 21:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.