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

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...

• Try to use the fact that the converse of Lagrange does not always hold, think some example about it. Commented Aug 13 at 13:59
• The answer to the second question is no in general. In fact, if it holds for all primes for a given group, then that group is solvable. See math.stackexchange.com/questions/661786/… Commented Aug 13 at 14:21
• "Can a nonabelian simple group have odd order?" is a simple question that does not involve any 'higher concepts' and can be understood by any students who has studied up to Sylow's Theorems... but any answer to that will involve difficult concepts (and the proof took an entire journal issue). Commented Aug 13 at 14:31
• If the answer to the first question were always true, then all the finite groups would be solvable. Start with any $G$. Let $p$ be the smallest prime factor of its order. There is a subgroup of order $p$, so by this property $G$ would have a subgroup $H$ of index $p$ also. But such a subgroup is normal (minimality of $p$ is crucial here). Then apply the same reasoning to $H$. Commented Aug 13 at 15:26

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 $$12/2 = 6$$.

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 $$n = p$$.

• What about my second question? Is that true? Commented Aug 13 at 14:13
• See the other answers/comments and the edit. Commented Aug 13 at 15:04

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 $$a$$.

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.

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).

• Why is $|G|$ even? Commented Aug 13 at 14:10
• @AvyakthaAchar It is a theorem of Thompson and Feit that a group of odd order is solvable, hence a nonabelian simple group cannot have odd order. Commented Aug 13 at 14:19
• I see, thank you. Commented Aug 13 at 14:21
• What about my second question? Commented Aug 13 at 14:21

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?