User:J D Bowen/For Julie

First Problem
Suppose $$a\in G \ $$ and $$|a|=n \ $$, so that $$a^n = e \ $$ and $$ a^m \neq e \ $$ when $$m<n \ $$.

If $$f:G\to H \ $$ is a homomorphism, then

$$f(a)^n = f(a^n) = f(e)=e \ $$, and so $$|f(a)|\leq n \ $$.

Suppose that $$|f(a)| \ $$ does not divide $$n \ $$. Then $$n = k|f(a)|+r \ $$, where $$r \ $$ is a remainder $$r<|f(a)|\leq n \ $$. Then

$$f(a)^n = f(a)^{k|f(a)|} f(a)^r = 1f(a)^r \neq e \ $$.

But $$f(a)^n = e \ $$. Hence $$|f(a)| \ $$ divides $$n \ $$.

Second Problem
Let $$R \ $$ be a nontrivial ring with unity, and $$a\in R \ $$ with $$a^2 = 0 \ $$.

Then $$(a+1)(a-1)=a^2-1 = -1 \ $$. Then $$(a+1)(1-a)=1 \ $$ and $$-(a+1)(a-1) = 1 \ $$, so these two elements are invertible.