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 $H = Ha$.

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    $\begingroup$ Yes, if $H\subset Ha$ then $H=Ha$, but you can't infer that on cardinality grounds, however there is an algebraic proof. $\endgroup$ – Lord Shark the Unknown Oct 29 '18 at 16:18
  • $\begingroup$ I understand that there is another solution, my question was why my method isn't correct. $\endgroup$ – Nerdlord Oct 29 '18 at 16:19
  • $\begingroup$ I think your teacher may refer to cases as $2\mathbb{Z}\leq \mathbb{Z}$? $\endgroup$ – Balloon Oct 29 '18 at 16:20
  • $\begingroup$ 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. $\endgroup$ – Nerdlord Oct 29 '18 at 16:26
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    $\begingroup$ 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. $\endgroup$ – user247327 Oct 29 '18 at 16:43

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