# Does a group of order $400$ always have a subgroup of order $200$?

Does a group of order $$400$$ always have a subgroup of order $$200$$?

I was considering some simple applications of Sylow theorems. I have made some questions. Among them, there was one which asks to prove that a group of order 200 always have a subgroup of order 100. This question has a simple solution since the Sylow group of order $$25$$ is unique and hence normal. After this, I became curious about the existence of normal subgroups of order $$2^k p^l$$ of groups of order $$2^{k+1} p^l$$.

The following two paragraphs are about my failed attempt which may not helpful.

For a group of order $$400$$, say $$G$$, let $$n_5$$ be the number of Sylow group of order $$25$$. If $$n_5 =1$$, we are done. So suppose that $$n_5 = 16$$. Let $$A$$ and $$B$$ be two Sylow subgroups of order $$25$$. Analysing the number of elements in the set $$AB$$, one can conclude that the order of $$A \cap B$$ to be $$5$$. This forced $$A \cap B$$ to be normal subgroup in $$A$$ and $$B$$. I was considered the normalizer $$N(A \cap B)$$. For this should contain $$A$$ and $$B$$, the order of $$N(A \cap B)$$ should be $$200$$ or $$400$$. If it was $$200$$, we are done. So suppose that for every pair of different Sylow groups of order $$25$$ $$A$$ and $$B$$, $$A \cap B$$ is normal. If there occur two different intersections of order $$5$$, we are done since the product of two such normal subgroups of order $$5$$ will provide a normal subgroup of order $$25$$.

The point I was stuck is there. Specifically, if a counterexample exists, it should have $$16$$ Sylow subgroups or order $$25$$ whose intersection is of order $$5$$.

The followings are just some consideration for generalization which may "not even wrong"

Most generally, I'm curious about the condition of $$n$$ which forces to exist a subgroup of order $$n$$ of a group of order $$2n$$. Note that as all of you know, there are some groups of even order without having subgroups of index $$2$$. All simple groups of even order have this property since a subgroup of index $$2$$ is automatically normal. Anyway, let me just state the general question.

Find a simple criterion for a positive integer $$n$$ to have the property that every group of order $$2n$$ has a subgroup of index $$2$$.

Finally, thank you for your attention.

• For the first part with Sylow compare with this post. – Dietrich Burde Jul 5 at 12:22
• Every group of order $\;2n\,,\,\,n\;$ odd, has a (normal, indeed) subgroup of order $\;n\;$ . If $\;n\;$ isn't even this can fail, as the example of $\;A_4\;$ shows. – DonAntonio Jul 5 at 12:47
• Many thanks for comments. I'm interested in the DonAntonio's. How can we argue to prove the statement? – seoneo Jul 5 at 14:29
• Do you want to prove that every group of order $2n$, where $n$ is odd, has a subgroup of index $2$? – the_fox Jul 5 at 14:39
• @the_fox Yes. I want. – seoneo Jul 5 at 14:49

There is a group $$H$$ of order 80 whose normal subgroups have orders 1,16,80. Taking $$G=\mathbb{Z}_5\times H$$ gives you a group of order 400 with no normal subgroup of index 2.

To get an explicit version of $$H$$, just take the group of matrices $$\left\{ \begin{pmatrix} \alpha & 0 \\u & \alpha^{-1} \end{pmatrix}\mid u\in GF(16), \alpha\in GF(16)^{*}, o(\alpha)|5 \right\}.$$

• GAP says that there are only two groups with no subgroup of index 2. One is your example and the other has shape $G = C_2^4 \rtimes C_{25}$. I wonder if it takes a lot of work to show this. – the_fox Jul 5 at 13:21
• I am a pre-GAP dinosaur, and I have no doubt that this is provable using elementary techniques. The fact that $5$ doesn't divide $2^n-1$ until $n=4$ helps enormously. – ancientmathematician Jul 5 at 13:24
• What a marvelous example! – seoneo Jul 9 at 17:16

Ancientmathematician has answered your first question. You have asked for a simple criterion to determine when a group of even order has a subgroup of index $$2$$. One such criterion is that the highest power of $$2$$ which divides $$|G|$$ is $$2^1$$, as mentioned in the comments.

Call $$n \in \mathbb{N}$$ a supersoluble number if every group of order $$n$$ is supersoluble. It is a standard theorem that a (finite) supersoluble group has subgroups of each possible order. In particular, if $$G$$ is a supersoluble group of even order then $$G$$ has a (necessarily normal) subgroup of index $$2$$.

A theorem which can act as a supplement to the "$$m=\text{odd}$$" criterion is the following one. First, define $$\psi$$ to be the multiplicative function defined on prime powers as $$\psi(p^k) = (p^k-1)(p^{k-1}-1)\dots(p-1)$$.

Theorem: Let $$n=p_1^{a_1}p_2^{a_2} \dots p_r^{a_r}$$ be a positive integer, where $$p_1 < p_2 < \dots < p_r$$. Then $$n$$ is a supersoluble number if and only if:

1. For all $$1 \leq i \leq r$$, the distinct prime factors of $$\gcd(n,\psi(p_i^{a_i}))$$ are the same as those of $$\gcd(n,p_i-1)$$.

2. If there exists $$i \neq k$$ such that $$p_i \leq a_k$$ (i.e. the value of some prime factor of $$n$$ is less than the multiplicity of another), then

(a) There does not exist a prime $$p_j$$ such that $$p_i | p_j-1$$ and $$p_j | p_k-1$$, and

(b) $$a_i \leq 2$$, and if $$a_i = 2$$, then $$p_i^2 | p_k - 1$$.

Applying this theorem to $$n=20$$ gives at once that $$20$$ is a supersoluble number, thus a fortiori every group of order $$20$$ has a subgroup of index $$2$$. Note here that the "$$m=\text{odd}$$" criterion tells you nothing about $$n=20$$. On the other hand, $$n=12$$ is not a supersoluble number because $$\gcd(12,\psi(2^2)) =3$$, whereas $$\gcd(12,1) =1$$. Indeed, $$A_4$$ is not supersoluble and it so happens that $$A_4$$ has no subgroup of index $$2$$.

Warning: if $$n$$ is not a supersoluble number, it does not follow that some group of order $$n$$ fails to have a subgroup of some order $$d \mid n$$ (much less specifically $$d=\frac{n}{2}$$). James' comment provides a concrete counterexample: every group of order $$224=2^5 \cdot 7$$ has subgroups of each possible order (i.e. $$224$$ is a Lagrangian number), but there are groups of order $$224$$ which are not supersoluble. I don't know if there are any useful arithmetic criteria guaranteeing that a number $$n$$ is Lagrangian.

Added: There is another criterion you can use which subsumes the "$$m=\text{odd}$$" one. Recall the definition of $$\psi$$ above. If $$G$$ is a group of order $$n$$ and if $$\gcd(n,\psi(2^a))=1$$, where $$2^a$$ is the highest $$2$$-power dividing $$n$$, then $$G$$ is $$2$$-nilpotent (see e.g. Isaacs' Finite Group Theory book, Corollary $$5.29$$). It is easy to see that a $$2$$-nilpotent group will always have a subgroup of index $$2$$.

• Yes, 224 is not a supersoluble number (there are two groups of order 224) that are not supersoluble, but it is a Lagrangian number (every group of that order satisfies the converse of Lagrange's theorem). It seems to be the only order up to 1000. Going up to 10000, there are 29 such orders, of which 27 are even. – James Jul 5 at 19:54
• Interesting. Cheers! – the_fox Jul 5 at 19:58
• There is a criterion for a number to be Lagrangian given in the paper, T.R. Berger, A Converse to Lagrange's Theorem, J. Austral. Math. Soc. 15 (Ser. A) 1978, 291-313. – James Jul 5 at 23:33
• I have had a look. That thing is formidable. – the_fox Jul 6 at 16:43
• I'm sorry. I'm somewhat new to using this site. I intended to select two posts as answers. But I have failed... – seoneo Jul 9 at 18:25

Let me also add a characterization for a group to have (or not to have) a subgroup of index $$2$$. One can show the following:

A group $$G$$ has no index $$2$$ subgroup iff $$G$$ is generated by squares, i.e. $$G = \langle \lbrace g^2 \mid g \in G\rbrace \rangle$$.

The group $$\langle \lbrace g^2 \mid g \in G\rbrace \rangle \subset G$$ is actually always a normal subgroup, but one can get some additional information if it has index $$2$$ in $$G$$:

A group $$G$$ has exactly one subgroup of index $$2$$ iff $$\langle \lbrace g^2 \mid g \in G\rbrace \rangle$$ has index $$2$$ in $$G$$.

• Thanks for a good point which I didn't know before! – seoneo Jul 9 at 22:20