Subgroup Problem

**GoldendoodleMom** · April 16th 2008, 05:47 AM

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...

**flyingsquirrel** · April 16th 2008, 06:13 AM

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

**GoldendoodleMom** · April 16th 2008, 06:27 AM

Originally Posted by flyingsquirrel

In fact there is only one operation possible : *

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

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

**Isomorphism** · April 16th 2008, 06:32 AM

Originally Posted by GoldendoodleMom

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

By data $x,y \in H$ , which means they have order 2. Saying x has order 2 implies x*x = e.

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