Proving Subgroups in Abelian Groups

[fishturtle1]

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.

 

[Orodruin]

Did you miss the statement that the group should be Abelian in the problem statement? It is just in your thread title. Always reproduce the problem statement verbatim.

So i just assumed the operator is multiplication.

This is an abstract group. The only thing you need to know are the group axioms and it really does not matter what you call your group operation or how you denote it.

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

This is a long way around. The identity is in H because it is its own inverse.

all elements' inverse is in H,

Even if it is obvious, you should state why this is the case. When you check if a subset is a subgroup, tick off the group axioms one by one and make the argument explicit (except associativity, the group operation is associative in H because it is associative in H, always).

 

[fishturtle1]

Did you miss the statement that the group should be Abelian in the problem statement? It is just in your thread title. Always reproduce the problem statement verbatim.This is an abstract group. The only thing you need to know are the group axioms and it really does not matter what you call your group operation or how you denote it.
This is a long way around. The identity is in H because it is its own inverse.Even if it is obvious, you should state why this is the case. When you check if a subset is a subgroup, tick off the group axioms one by one and make the argument explicit (except associativity, the group operation is associative in H because it is associative in H, always).

Sorry I meant to say "Let G be an abelian group" where i said "Let G be a group".

to your second point: In the book, in the opening paragraph it says "Let G be a group and S a nonempty subset of G. It may happen (though it doesn't have to) that the product of every pair of elements of S is in S. If it happens, we say hat S is closed with respect to multiplication. Then, it may happen that the inverse of every element of S is in S. In that case, we say that S is closed with respect to inverses. If both these things happen, we call S a subgroup of G."

I take this to mean that for H to be a subgroup, it must be closed under multiplication, so shouldn't the operation of H have to be multiplication?

to your third point: Ok i think i get it, because for all x in H, $x^{-1}$ is also in H, so H is closed under inverses which implies
e $\epsilon$ H.

to your fourth point: by stating why, could I just say "for all x in H, $x^{-1}$ must also be in H so H is closed under inverses"?

 

[Orodruin]

I take this to mean that for H to be a subgroup, it must be closed under multiplication, so shouldn't the operation of H have to be multiplication

As I said, it does not matter what the group operation is. Sometimes the group operation of an abstract group is generically referred to as "group multiplication" or just "multiplication". It is up to you to define what this means for a particular group.

Regarding the fourth point. It is much more trivial than that since $x^{-1}=x$ in this case. You want a statement to the effect "if $x\in H$ then $x^{-1}\in H$ because ... (fill in the dors)

 

[fishturtle1]

As I said, it does not matter what the group operation is. Sometimes the group operation of an abstract group is generically referred to as "group multiplication" or just "multiplication". It is up to you to define what this means for a particular group.

Regarding the fourth point. It is much more trivial than that since $x^{-1}=x$ in this case. You want a statement to the effect "if $x\in H$ then $x^{-1}\in H$ because ... (fill in the dors)

Ok, so for example, in my first post, when I said xy = $x^{-1}y^{-1}$, I shouldn't think of multiplication of numbers or matrices, I should just think of some operation between x and y and that same operation between $x^{-1}$ and $y^{-1}$.

if $x\in H$ then $x^{-1}\in H$ because..$x = x^{-1}$?

 

[Orodruin]

Ok, so for example, in my first post, when I said xy = $x^{-1}y^{-1}$, I shouldn't think of multiplication of numbers or matrices, I should just think of some operation between x and y and that same operation between $x^{-1}$ and $y^{-1}$.

if $x\in H$ then $x^{-1}\in H$ because..$x = x^{-1}$?

Right. A priori, we do not even need to know what the group elements are, they could be fruits (although at least I would struggle with defining a meaningful group operation). What matters is that the group axioms are satisfied, here with the additional requirement of the group being Abelian.

I would write it the other way around ($x^{-1} = x \in H$), but the meaning is the same.

 

[fishturtle1]

Right. A priori, we do not even need to know what the group elements are, they could be fruits (although at least I would struggle with defining a meaningful group operation). What matters is that the group axioms are satisfied, here with the additional requirement of the group being Abelian.

I would write it the other way around ($x^{-1} = x \in H$), but the meaning is the same.

Understood, thank you for clearing up my confusion

 

[WWGD]

Wow, same advice, different problem: If G is a group and $S \subset G$ then S is a subgroup if for all $x,y \in S , xy^{-1} \in S$