isomorphism and normal subgroup?

2016-11-07 8:56 pm

Let G be a group and let D = {(g, g) : g ∈ G}.
(a) Show that D is isomorphic to the group G.
(b) Show that D is a normal subgroup of the group G×G if and only if G is an abelian group.

Can I write Let G isomorphic to D, f:G-> D defined by f(g) = (g,g)?
How to prove onto?

回答 (1)

[匿名]

2016-11-07 9:12 pm

✔ 最佳答案

Yes, that would be one isomorphism. There may be others.

Isn't it trivial to show that it's onto? Let (x,x) be any element of D. Then f(x) = (x,x), so so (x,x) is in the range of f.

Suppose D is normal in GxG. Then (x, y)(d, d)(x^-1, y^-1) = (xdx^-1, ydy^-1) is in D for for any d, x and y in G.
Since that is in D, you must have
xdx^-1 = ydy^-1
Let y = e, and you get
xdx^-1 = d
xd = dx, so G must be Abelian.

Going the other direction:
IF G is Abelian, so is GxG, and all subgroups are normal.

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