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.

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- Mar 9th 2013, 11:25 AMlovesmathSubgroups
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.

- Mar 9th 2013, 12:24 PMLockdownRe: 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, for all g in G. In particular, is either e or a.

Note that is an isomorphism (it is an automorphism, actually). This means that it must preserve the order of the element.

Hence, , or else we would have a contradiction.

Thus, 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. - Mar 9th 2013, 12:29 PMILikeSerenaRe: Subgroups
- Mar 10th 2013, 08:51 AMlovesmathRe: 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).

- Mar 10th 2013, 10:54 AMILikeSerenaRe: Subgroups
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.

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and every element in the center commutes with all other elements.

Good!

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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.

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Thus, H is a subgroup of Z(G).