1. ## Subgroups

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.

2. ## Re: Subgroups

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, $gag^{-1} \in H$ for all g in G. In particular, $gag^{-1}$ is either e or a.
Note that $a \mapsto gag^{-1}$ is an isomorphism (it is an automorphism, actually). This means that it must preserve the order of the element.
Hence, $gag^{-1}=a$, or else we would have a contradiction.

Thus, $gag^{-1} = a \Leftrightarrow ga=ag$ 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.

3. ## Re: Subgroups

Originally Posted by lovesmath
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.
Hi lovesmath!

If |H|=2, then we can write H={e,h}, which we know is a normal subgroup of G.

Is e an element of Z(G)?
Is h?
What are the requirements to call a set a subgroup? Are they satisfied in this case?

4. ## Re: Subgroups

Both e and h are elements of Z(G), and every element in the center commutes with all other elements. So, e and h commute with each other since they are the only elements in H. Thus, H is a subgroup of Z(G).

5. ## Re: Subgroups

You're mixing things up a bit.

Originally Posted by lovesmath
Both e and h are elements of Z(G),
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.

and every element in the center commutes with all other elements.
Yes that is the definition of Z(G).
Good!

So, e and h commute with each other since they are the only elements in H.
Yes, e and h commute with each other.
Not because they are the only elements in H, but simply because they are elements of a normal subgroup.

Thus, H is a subgroup of Z(G).
I'm afraid this does not follow from the previous.