# Confusion about infinite order groups, subgroups and comparing order between them.

I was given feedback for a homework problem,

Let $$G$$ be a group. Let $$H$$ be a subgroup of $$G$$. Let $$a \in G$$. Let $$Ha$$ be a subgroup of $$G$$.

The problem was to show that $$Ha = H$$.

I established that $$H \subset Ha$$ and then I made the argument that since $$Ha = \{ha \vert h \in H\}$$, $$\vert Ha \vert \leq \vert H \vert$$ so $$H = Ha$$

I was marked wrong and given the justification that $$\vert Ha \vert \leq \vert H \vert$$ isn't valid if $$\vert H \vert = \infty$$

I wanted to understand why, as I think the point still holds that $$Ha$$ can have at most as many elements as $$H$$, so if $$H$$ is a subset of $$Ha$$ then $$H = Ha$$.

• Yes, if $H\subset Ha$ then $H=Ha$, but you can't infer that on cardinality grounds, however there is an algebraic proof. – Lord Shark the Unknown Oct 29 '18 at 16:18
• I understand that there is another solution, my question was why my method isn't correct. – Nerdlord Oct 29 '18 at 16:19
• I think your teacher may refer to cases as $2\mathbb{Z}\leq \mathbb{Z}$? – Balloon Oct 29 '18 at 16:20
• Wouldn't it be fair to say that $\vert 2Z \vert \leq \vert Z \vert$ as every element of $2Z$ is generated by a multiplication between a number in $Z$ and $2$? That's what I don't understand. – Nerdlord Oct 29 '18 at 16:26
• Z is the set of all integers and 2Z is the set of all even integers. Obviously they are NOT the same set but have the same cardinality- they are both countably infinite. – user247327 Oct 29 '18 at 16:43