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.

Third Problem
Let $$H \triangleleft G $$. We aim to show $$G/H \ $$ is cyclic if and only if $$\exists a \in G : \forall x \in G, \ \exists n\in \mathbb{N}: xa^n \in H \ $$.

The hint given is that $$Ha=Hb \iff ab^{-1}\in H \ $$ and $$Ha=H \iff a \in H \ $$.

To show the $$\Leftarrow \ $$, assume $$\exists a \in G : \forall x \in G \exists n\in \mathbb{N}: xa^n \in H \ $$.

Consider the natural surjection homomorphism $$\phi(x)=\overline{x} \ $$.

Then $$\phi(ax^n) \in \phi(H) = \left\{{ 1 }\right\} \ $$, but we also have

$$\phi(xa^n)=\phi(x)\phi(a^n) \ $$.

So $$\phi(x)^{-1} = \phi(a)^n \ $$.

But since $$\forall y \in G/H, \exists x\in G : \phi(x)^{-1}=y \ $$, we have for all $$y \in G/H, \exists n \in \mathbb{N} \ $$ such that $$ \overline{a}^n =y \ $$.

Hence $$G/H = \langle \overline{a} \rangle \ $$ and so $$G/H \ $$ is cyclic.

Now we aim to show $$\implies \ $$, so assume $$G/H \ $$ is cyclic, and call the generator $$y \ $$. Then for all $$z \in G/H, \exists n\in\mathbb{N}: y^n=z^{-1} \ $$, which implies $$zy^n=1 \ $$. Let $$x\in z, \ a\in y \ $$. Then $$xa^n \ $$ is in the preimage of $$1 \ $$, which is precisely $$H \ $$.