# Let $(G,\cdot)$ be a group of order $2n$ with $n$ elements of order $2$. Prove $n$ is odd and $G$ has an abelian subgroup of order $n$.

Let $$(G,\cdot)$$ be a group of order $$2n$$ which has exactly $$n$$ elements of order $$2$$. Prove that $$n$$ is odd and that $$G$$ has an abelian subgroup of order $$n$$.

For the first part, the elements whose order is greater than $$2$$ can be grouped in pairs of the form $$\{x,x^{-1}\}$$. Now it easily follows that $$n=\text{even}-\text{odd}=\text{odd}$$.

For the second part, I though about considering the set $$H=\{x_1,x_2,\dots ,x_n\}$$ where $$x_i$$ is an element of order $$2$$ $$\forall i=\overline{1,n}$$.

I observed that for $$i\neq j$$ we have that $$x_i x_j \in G\setminus H$$, because otherwise $$\{e,x_i,x_j, x_i x_j\}$$ would be a subgroup of order $$4$$ of $$G$$, which would contradict Lagrange's theorem. I didn't know how to proceed from here.

• Can you use this post? Commented Jan 18, 2020 at 20:32
• @Dietrich Burde It definitely is useful, but how should I prove that $G \setminus H$ is a subgroup(in my notations)? Commented Jan 18, 2020 at 20:41

I will try to keep to the spirit of your approach. Let's write $$K$$ for your $$G \setminus H$$, the set of elements of order other than $$2$$. It's a good bet that $$K$$ should be such a subgroup, since the whole question makes it sound like your group behaves sort of like the dihedral group of order $$2n$$, where the $$n$$ reflections have order $$2$$ and the $$n$$ rotations are a cyclic subgroup.

Note that $$x_1 x_1, x_2 x_1, \dotsc, x_n x_1$$ are all distinct, since $$x_i x_1 = x_j x_1 \implies x_i = x_j$$. Therefore each element of $$K$$ can be written as $$x_i x_1$$ for some $$i$$. Then we can use the efficient subgroup criterion: if $$a = x_i x_1$$, $$b = x_j x_1$$ are two arbitrary elements of $$K$$, then $$ab^{-1} = x_i x_1 x_1 x_j = x_i x_j \in K$$, and also $$e \in K$$, so $$K$$ is a subgroup.

Then by this question, we're done. I've copied over the important bit for completeness:

Let $$a, b \in K$$ be arbitrary and $$x_1$$ have order 2. Since $$x_1 \notin K$$, $$x_1 K = H$$, ie $$x_1 a$$ has order 2, so $$x_1 a x_1 a = e \implies x_1 a x_1 = a^{-1}$$. Then $$x_1 a^{-1}b^{-1} x_1 = (a^{-1}b^{-1})^{-1} = ba$$ (since $$a^{-1}, b^{-1} \in K$$), but also $$x_1 a^{-1}b^{-1} x_1 = x_1 a^{-1} x_1 x_1 b^{-1} x_1 = ab$$, so $$K$$ must be abelian.

PS: It wasn't really necessary to use the efficient subgroup criterion, it would have worked just as well to say that for any $$i$$, $$x_i x_j = x_i x_k \implies x_j = x_k$$ and $$x_j x_i = x_k x_i \implies x_j = x_k$$ and directly deduce closure and inverses. It's just that that becomes a little messy.

• thank you very much ! I had also observed that $x_1 x_i$, $\forall i =\overline{1,n}$ are all distinct and are part of $K$, but I hadn't realised it was so easy to conclude that it is a subgroup. Commented Jan 18, 2020 at 22:39
• I think it is good to copy a duplicate as a crucial part, since there has been spend a lot and time and effort for the answers. Why should it be ignored? Commented Jan 19, 2020 at 9:26

Suppose that $$|G| = 2n$$ and that the number of elements of order $$2$$ in $$G$$ is $$n$$.

You can see that the number of $$x \in G$$ such that $$x^2 = 1$$ is even (for a proof, partition elements of $$G$$ into sets $$\{x,x^{-1}\}$$). This implies that the number of elements of order $$2$$ in $$G$$ is odd, so $$n$$ is odd.

Now consider the regular permutation representation of $$G$$, that is, $$G$$ acting on itself by left multiplication). In this action, an element $$x \in G$$ of order $$2$$ acts with cycle structure $$(x_1 x_2) (x_3 x_4) \cdots (x_{2n-1} x_{2n}).$$

In particular $$x$$ acts as an odd permutation, so $$G$$ has a normal subgroup $$H$$ of index $$2$$. (In general if $$G \leq S_n$$ and $$G \not\leq A_n$$, then $$G \cap A_n$$ is a normal subgroup of $$G$$ with index $$2$$.)

Let $$x \in G$$ have order $$2$$ and let $$H \leq G$$ be a subgroup of index $$2$$. Now $$H$$ has no elements of order $$2$$ since $$|H| = n$$ is odd, so every element of $$xH$$ has order $$2$$.

Then for all $$h \in H$$, we have $$xhx^{-1} = xhx = (xh)^2h^{-1} = h^{-1}.$$ So conjugation by $$x$$ gives an automorphism $$h \mapsto h^{-1}$$ of $$H$$. The inverse map is an automorphism if and only if the group is abelian, so $$H$$ must be abelian.