0
$\begingroup$

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 $G$.

$\endgroup$
  • 3
    $\begingroup$ $A$ is not necessarily a subgroup of $G$, so it would be quite remarkable if that were "easy to show" $\endgroup$ – Omnomnomnom Jan 22 at 10:20
8
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.