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 \ $.