- #1

fishturtle1

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## Homework Statement

Let G be a group. if H = ##{x \epsilon G : x = x^{-1}}##, that is H consists of all elements of G which are their own inverses, prove that H is a subgroup of G.

## Homework Equations

to show H is a subgroup of G, show that H is closed under the operation of G and every element in H has its inverse in H.

## The Attempt at a Solution

I'm kind of confused because there is no operator for G specified. Like in previous examples on abelian groups, the book defined an operator. So i just assumed the operator is multiplication.

**EDIT: reread again...and realized they say a subgroup is closed under inverses and closed under multiplication. so my question is, is there anyway to make this proof clearer, or is it clear enough ?**

if x, y ##\epsilon## H, then ##x = x^{-1}## and ##y = y^{-1}##

xy = ##x^{-1}y^{-1}##

xy = ##(xy)^{-1}##

so H is closed under multiplication.

For all x##\epsilon##H, ##x^{-1} \epsilon## H

**EDIT: I think I could have stopped here, and said since H is closed under multiplication and closed under inverses it is a subgroup of G.**

since H is closed under multiplication, the identity element must also be in H because of this:

if x##\epsilon## H then ##xx^{-1}## = e ##\epsilon## H

H is closed under multiplication, all elements' inverse is in H, and the identity element is in H, so H is a subgroup of G.

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