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April 18th, 2012, 07:04 AM   #1
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Abstract Algebra normal groups. Quick please.

Let G be a group and H normal with G. Prove that if |H| = 2, then H is a subgroup of Z(G).
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April 18th, 2012, 08:54 AM   #2
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Re: Abstract Algebra normal groups. Quick please.

Here's what i have come up with. Is this right?

Since H is a subgroup, it must contain the identity, call it e. Call the non-identity element of H h. Thus, H = {h, e}. Since H contains its own inverses, h^2 = e (if h^2 = h, then h would have to be the identity).

Anyway, by the normality of H, we know that for any element g of G, gH = Hg. That is to say,
{gh, ge} = {hg, eg}, which means that
{gh, g} = {hg, g}

since h is not the identity, we know that gh?g and hg?g. Thus, in order for Hg and gH to be equal, we must have that gh = hg for an arbitrary element g. Thus, we have shown that H is a subgroup of Z(G).
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April 18th, 2012, 01:13 PM   #3
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Re: Abstract Algebra normal groups. Quick please.

your argument is correct.

here is another:

consider ghg^-1 for any element g in G, where h is the non-identity element of H.

since H is normal, ghg^-1 is in H. thus, either:

ghg^-1 = e --> gh = g --> h = e, contradiction! thus, this never happens.

or:

ghg^-1 = h --> gh = hg, for ALL g in G, hence h is in Z(G). since e is always in any subgroup, {e,h} = H is contained in Z(G).
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