# How much elements has a subset of a group with even elements

Can someone help me on this problem? Thank you so much!

Let $$G$$ a group with even number of elements and $$A$$ a subset such that $$A = \{ g \in G | g^2 = e \}$$. Show that $$A$$ has an even number of elements.

My try:

Hence $$G$$ has even elements that means $$G$$ has elements with order 2. Let $$x \in G$$ such that $$x^2 = e$$ (by Cauchy) then $$x = x^{-1}$$.. The same thing for $$x \in A$$. It's easy to show $$A$$ is a subgroup of $$G$$.

• $A$ is not necessarily a subgroup of $G$, so it would be quite remarkable if that were "easy to show" – Omnomnomnom Jan 22 at 10:20

The map $$x\mapsto x^{-1}$$ is an involution $$G\setminus A\to G\setminus A$$ without fixed points, therefore $$\lvert G\setminus A\rvert$$ is even. This proves that $$\lvert A\rvert$$ is even.
Remark: Proving that "$$A$$ is a subgroup of $$G$$" is either very easy or very hard, depending on one's propension towards drawing false conclusions: it's actually false in general.
I've got another interesting solution in time. By Cauchy theorem exists elements in $$A$$ with $$ord(x) = 2$$, $$x \in G$$. Let $$y \in G$$ be an element such that $$ord(y) > 2$$ so $$y$$ and $$y^{-1}$$ can be coupled in groups $$(y, y^{-1})$$ so hence $$G$$ has even number of elements and number of elements such that $$ord(y) > 2$$ is even we get $$A$$ has even number of elements.