1. ## Subgroup Problem

Let G be an abelian group. Let H be the subset of G consisting of the identity e together with all elements of G of order 2. Show that H is a subgroup of G.

I'm not sure how to go about showing closure. Can it be assumed that any operation on elements of order 2 will produce another element of order 2? Doesn't seem right...

2. Originally Posted by GoldendoodleMom
Can it be assumed that any operation on elements of order 2 will produce another element of order 2?
In fact there is only one operation possible : *

For the closure : let $\displaystyle x,\,y \in H$,
$\displaystyle (x*y)^2=(x*y)*(x*y)=x*y*x*y=x*x*y*y$ thanks to associativity and commutativity
Then, $\displaystyle x*x*y*y=(x*x)*(y*y)=e*e=e$ because $\displaystyle x$ and $\displaystyle y$ have order 2
Hence, $\displaystyle x*y$ also has order 2 and lies in $\displaystyle H$.

3. Originally Posted by flyingsquirrel
In fact there is only one operation possible : *

Then, $\displaystyle x*x*y*y=(x*x)*(y*y)=e*e=e$ because $\displaystyle x$ and $\displaystyle y$ are of order 2
Hence, $\displaystyle x*y$ is of order 2 and lies in $\displaystyle H$.

Thanks for your help flyingsquirrel. I'm just stuck on why x*x or y*y = e?

4. Originally Posted by GoldendoodleMom
Thanks for your help flyingsquirrel. I'm just stuck on why x*x or y*y = e?
By data $\displaystyle x,y \in H$, which means they have order 2. Saying x has order 2 implies x*x = e.