# 1 maximal subgroup -> prime order

1. Dec 6, 2009

### boboYO

'Prove that if a finite group G has only one maximal subgroup M, then |G| is the power of a prime'

I've somehow deduced that no finite group has only one maximal subgroup, and I'm having trouble seeing where I went wrong.
This is what I have:

Let $$H_1$$ be a subgroup of G. Either $$H_1$$ is maximal and equal to M or it is not maximal and there is a $$H_2$$ such that $$H_1<H_2<G$$ (using < to mean proper subgroup). Apply the same argument to $$H_2$$ and we get an ascending chain of subgroups. Since G is finite the process must end eventually (when we reach M).

Thus every subgroup of G is a subgroup of M and hence every element of G is an element of M, and G=M, a contradiction.

So would someone like to point out the flaw in the above reasoning? thanks.

Last edited: Dec 6, 2009
2. Dec 6, 2009

### Hurkyl

Staff Emeritus
Try substituting for G a particular group where the conclusion is obviously wrong. Do you know any groups with only one maximal subgroup?

The most obvious ones to me to try are the two-element group, or maybe the cyclic group on 4 elements.

3. Dec 6, 2009

### boboYO

Thanks Hurkyl. That should've been the first thing to do. For anyone interested, the problem is with the last sentence. It does not follow that every element of G is an element of M: the elements of G that generate G do not have to be in M.

4. Sep 10, 2011

### Borvorn

I know that this post is old, but since it is top result when i google this problem, I think it would be good if i give further detail of solution.

Let $a \in G \setminus M$. Then $a$ generate $G$ which mean $G$ is cyclic. Suppose $|G| = p_{1}^{r_{1}}...p_{n}^{r_{n}}$ where $p_{i}$ is prime. So $a^{p_{i}}$ and $a^{p_{j}}$ generated a different maximal subgroup which is contradiction.

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