If there is a subgroup of order $d$, then is there a subgroup of order $n/d$?

Question

Let $G$ be a finite group of order $n$ and let $H$ be a subgroup of $G$ of order $d$ (Lagrange theorem tells us that $d\mid n$). My question is whether there exists a subgroup $H'$ such that order of $H'$ is $n/d$ (maybe we can call $H'$ a complement of $H$).

Hoping that this result is true, I attempted to prove this result by trying to construct a group action of G on some set X such that the size of the orbit of some element is $d$. Then by orbit stabiliser theorem, the order of the stabiliser would be $n/d$. But I wasn't able to construct such a group action in general.

Is the result true or is there any counter example? I would be grateful for any help.

In case the above result is not true, at least is it true that if $|G|=p_1^{k_1}\cdots p_n^{k_n}$, for distinct primes $p_1,\cdots, p_n$, then there is a subgroup $H_i$ of order $|G|/p_i^{k_i}$ (something like a complement of a Sylow $p_i$-subgroup) for each (or) for some $i \in \{1,\cdots,n\}$?

Do any of the above two questions hold in any special cases?

I know only basic group theory upto Sylow's Theorems, and answers without involving higher concepts like solvability would be preferable...

Answer 1

The answer is no.

For example, the alternating group $G = A_4$ is of order $12$. It has a subgroup of order $2$, but it has no subgroup of order $12/2 = 6$.

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Edit: As others have noted in comments/answers, your second question also has a negative answer.

I believe that for $G = S_n$ with $n \geq 5$ it fails almost always.

Specifically, suppose that $n \geq 5$, and let $p$ be prime divisor of $|G| = n!$. Let $p^k$ be the order of a $p$-Sylow in $G$. Then $G$ does not have a subgroup of order $|G|/p^k$, unless $k = 1$ and $n = p$.

Answer 2

There are plenty of examples that your first question has a negative answer.

The second question likewise has a negative answer in general (but holds for solvable groups). A counterexample to the second question is $S_5$, of order $120=5\times 3\times 8$; it has a subgroup of order $8$ (its Sylow $2$-subgroup), but no subgroups of order $15$: a group of order $15$ must be cyclic, but $S_5$ has no elements of order $15$. Same with $A_5$, which has subgroups of order $4$ but no subgroup of index $4$.

The second question does hold for solvable groups, by Hall's Theorem. Hall's Theorem is a generalization of Sylow's Theorem, and states among other things that a finite group $G$ is solvable if and only if for any positive integers $a$ and $b$ such that $|G|=ab$ and $\gcd(a,b)=1$, $G$ has a subgroup of order $a$.

Answer 3

As an answer to your second question, there is a theorem known as the Schur-Zassenhaus Theorem. It states the following:

Let $G$ be a finite group and $N$ a normal subgroup whose order is coprime to the order of $G / N$. Then $G$ is the semidirect product of $N$ and $G/N$.

So if $P$ is a normal Sylow $p$-subgroup, it seems like the answer to your question is a yes. One can say even more if you know this holds for all primes.

Answer 4

Let $G$ be any noncyclic finite simple group. Then $|G|$ is even, hence $G$ has an element (and subgroup) of order $2$, but it has no subgroup of index $2$ (because that would be a normal subgroup).

Answer 5

You should also check out CLT (Converse Langrange Theorem) groups, see for example If a group contains a subgroup with the order of each of its divisors, is it abelian?