So I want to prove that every subgroup of a finite group is contained in a maximal subgroup.

Let $G$ be a finite group, then we can say that $|G| = n \in \mathbb{N}$. Now let $H_1 < G$ If there exists no $H_2 < G: H_1 < H_2 < G$ then we are done since $H_1$ is maximal and is contained in itself. This should be ok, right?

In the case where such a $H_2$ exists, we have $H_1 < H_2 < G$. If $H_2$ is maximal we are done and so on...

By Lagrange theorem we have that $|H_1|$ divides $|H_2|$ and $|H_2|$ divides $|G|$. (in the general setting of a chain of proper subgroups we have that $|H_i|$ divides $|H_{i+1}|$)

But $|G| =n$ has a unique and distinct prime factorization which implies that we can not have a infinite chain of proper subgroups: $H_1 < H_2 < H_3 < ... < G$ between some $H_i$ and $G$. Since $|H_{i+1}| > |H_{i}|$ for every subgroup in the chain. That is, we can not have a infinite number of divisors of $n$ so there must exists a maximal $H_r < G$ in which our $H_1$ is contained (and all other $H_i$ in the chain).

This is messy and not very rigorous but is it plain out wrong or am I on the right track? Any help or corrections would be highly appreciated.

  • 2
    $\begingroup$ I think your proof is fine but if you want a more elegant argument you can try to consider the a subgroup which is not contained in a maximal subgroup with the maximum number of elements and try to get a contradiction. $\endgroup$ – Quimey Feb 7 '14 at 23:09

That is right, and rigorous, although there are clearer ways of writing it - and you don't actually need to use lagrange.

I'm presuming by maximal subgroup you mean a proper subgroup (i.e. not $G$) that is not contained in any other proper subgroup? (and also assuming that you only mean the result to hold for proper subgroups - else you need have $G$ as the only maximal subgroup, and clearly all subgroups are contained in $G$).

Clearer phrasing could be:

Suppose the statement is false. Then have some subgroups of $G$ that are not contained in any maximal subgroup. Note that the order of subgroups of $G$ is bounded by $n$. Then there must be some subgroup $H < G$ of largest order s.t. it is not contained in any maximal subgroup.

Then $H$ is not maximal, so $\exists H_1 \not = H$ a proper subgroup of $G$ with $H < H_1$. Since $\lvert H_1 \rvert > \lvert H \rvert$, $H_1$ is contained in some maximal subgroup $M$. But then $H < M$ - contradiction.


Let $H$ be a proper subgroup of $G$. Either $H$ is maximal, or $H < H_1$ with $H_1$ a proper subgroup of $G$ that has strictly larger order. Continuing in this way, we construct a chain of proper subgroups of $G$; $H < H_1 < H_2 < ...$. This gives a strictly increasing chain of integers $\lvert H \rvert < \lvert H_1 \rvert < \lvert H_2 \rvert < ...$ which is bounded above by $n$, hence must terminate - giving a maximal subgroup containing $H$.


Hint: this has nothing to do with groups. If $H$ is maximal, you have nothing to do. If it isn't, pick $H_1>H$. Repeat. Can you get an infinite chain? No, but not because of Lagrange's theorem.

In other words, your proof is almost fine, but you're using too much.


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