Group theory?

2015-11-09 4:25 am

Let G be a finite abelian group of order k and let n be a natural number
with gcd(n, k) = 1. Show that the map f : G → G defined by f(g) = g^n
is an isomorphism from G to G. Show that if gcd(n, k) is not equal to 1, then f
need not be an isomorphism.

In the first part, I write since f(gh)=(gh)^n=g^nh^n=f(g)f(h), the map f : G → G defined by f(g) = g^n
is an isomorphism from G to G.

how to prove the second part?

回答 (1)

[匿名]

2015-11-09 4:52 am

✔ 最佳答案

You haven't completely proven the first part. You've only shown f is a homomorphism. You also need to show it's 1-to-1.

For the second part, you know g^k = e for all g in G, since the order of every element must divide the order of the group. If gcd(k,e) = a>1, then some prime p divides a, and also divides k.

The fundamental theorem of Abelian groups says that G = product of subgroups of cyclic groups of prime power order. Once of the groups has order = some power of p, That group has an element h, not the identity, of order p. Therefore h^p = e and h^a = e and h^n = e. But e^n = e. So we have two elements of G which f maps to e, so it can't be 1-to-1.

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