Let G be a group and H be a normal subgroup of G. If |H|=2, then show that H<=Z(G). I know I have to show that H is a subgroup of the center of G, but I don't know what to do.
The order of H is 2, so it has two elements. Because H is a normal subgroup, it is a subgroup in particular, hence it must contain the identity element.
Hence, H = {e,a} where a is some element of order 2 (there can only be one element of order 1, and the order of the non-trivial element must divide 2, hence it must be 2).
Now since H is a normal subgroup, by definition, $\displaystyle gag^{-1} \in H$ for all g in G. In particular, $\displaystyle gag^{-1}$ is either e or a.
Note that $\displaystyle a \mapsto gag^{-1}$ is an isomorphism (it is an automorphism, actually). This means that it must preserve the order of the element.
Hence, $\displaystyle gag^{-1}=a$, or else we would have a contradiction.
Thus, $\displaystyle gag^{-1} = a \Leftrightarrow ga=ag$ for all g in G.
Hence, a commutes with every element in G. Since e does this too, H is a subset of the center of G.
Clearly, since a has order 2, this is a subgroup of the center of G.
You're mixing things up a bit.
You don't know yet whether both e and h are elements of Z(G).
That's what we're trying to find out.
You're not supposed to simply state it.
Yes that is the definition of Z(G).and every element in the center commutes with all other elements.
Good!
Yes, e and h commute with each other.So, e and h commute with each other since they are the only elements in H.
Not because they are the only elements in H, but simply because they are elements of a normal subgroup.
I'm afraid this does not follow from the previous.Thus, H is a subgroup of Z(G).