Let G be an abelian group.
If H = { x e G : x = x^-1 }, that is, H consists of all the elements of G which are their own inverses. Prove that H is a subgroup of G.
you need to show three things: H is nonempty (some texts require that you check that the identity be in the subgroup...which has to happen anyway, and still fulfills the nonempty condition), and closed under products and closed under inverses. 2 are obvious. where did you begin? where are you stuck?
Yes! the identity is its own inverse. so it is in H by definition. (we want it to be in H, not G. i made a mistake in my last post. H is what we want to show is a subgroup)
the next easiest condition to me is closure under inverses. do you know what that means? try to show it
it means, that if $\displaystyle \displaystyle g \in G$, then $\displaystyle \displaystyle g^{-1} \in G$. So yes, for each element in the set, its inverse will be in the set. if this happens, we say the set is "closed under inverses"
So now, is this true for $\displaystyle \displaystyle H$? If $\displaystyle \displaystyle x \in H$, is it true that $\displaystyle \displaystyle x^{-1} \in H$ ?
geez, have some self confidence, man! yes, that's right. if x is in H, then so will its inverse be in H, since x IS its own inverse.
now for the last condition. This one is harder, i will do it. First note, that $\displaystyle \displaystyle x = x^{-1}$ is the same as saying $\displaystyle \displaystyle x^2 = xx = e$. also note, that $\displaystyle \displaystyle G$ is Abelian if the operation on it is commutative. That is, $\displaystyle \displaystyle xy = yx$ for all $\displaystyle \displaystyle x,y \in G$. Now on to the proof.
We show that $\displaystyle \displaystyle H$ is closed under products (that is, if $\displaystyle \displaystyle x \in H \text{ and }y \in H$, then $\displaystyle \displaystyle xy \in H$). We need to show that $\displaystyle \displaystyle x,y \in H \implies xy \in H$, we will do this by showing that $\displaystyle \displaystyle xy = (xy)^{-1}$, provided $\displaystyle \displaystyle x$ and $\displaystyle \displaystyle y$ are in $\displaystyle \displaystyle H$.
Assume $\displaystyle \displaystyle x,y \in H$. Since $\displaystyle \displaystyle x,y \in G$, and $\displaystyle \displaystyle G$ is Abelian, we have $\displaystyle \displaystyle xy = yx$.
$\displaystyle \displaystyle \Rightarrow xyx = yxx$ .........multiplied both sides by x
$\displaystyle \displaystyle \Rightarrow xyx = y$ ............since $\displaystyle \displaystyle xx = e$, because $\displaystyle \displaystyle x \in H$
$\displaystyle \displaystyle \Rightarrow xyxy = yy$..........muliplied both sides by y
$\displaystyle \displaystyle \Rightarrow xyxy = e$ ..........since $\displaystyle \displaystyle yy = e$, because $\displaystyle \displaystyle y \in H$.
$\displaystyle \displaystyle \Rightarrow (xy)^2 = e$, as desired.
This completes the proof.
Group is just 3 things right?
Associativity, identity, and inverse
Abelian Group has 4 - Commutativity and the above 3. Correct?
1 last thing: Can give you give me a pointer on how to show that e is an element of H if H = {x e G : x^n = e} (n is a fixed integer)