If we have a group $H$ with $h\in H$, how do I prove that $|h|\ge |h^2|$? If we have a group $H$ with $h$ being an element in $H$, how would I prove that $|h|$ is greater than or equal to $|h^2|$?
I've looked through my notes to try and help me do this but didn't find anything useful and I'm quite confused to be honest. Help is needed and appreciated. 
Sorry about my notation, I'm still navigating myself around the website and haven't quite got to grips with this yet. 
 A: Let $h\in H$ with $\lvert h\rvert=n$. Consider 
$$\begin{align}
(h^2)^n&=h^{2n}\\
&=(h^n)^2\\
&=e^2\\
&=e.
\end{align}$$
Thus the order of $h^2$ is at most $n$, since, by definition, the order of $h^2$ is the smallest $m$ such that $(h^2)^m=e$.

Consider, for example, the group $\Bbb Z_4$ given by the presentation $$\langle a\mid a^4\rangle.$$ In this case $\lvert a\rvert=4$ whereas $(a^2)^2=e,$ so $\lvert a^2\rvert =2\le 4=\lvert a\rvert$.
Another example is the group $\Bbb Z$ given by the presentation $$\langle z\mid\varnothing \rangle.$$ Here both $\lvert z\rvert$ and $\lvert z^2\rvert$ are $\aleph_0$.
A: Lets prove this using a different view: The order of an element $h\in H$ is equal to the order of the minimal subgroup of $H$ containing $h$, denoted $\langle h\rangle$. That is, $|h|=|\langle h\rangle|$. This is a useful view in general, as for example it allows us to apply Lagrange's Theorem to prove that the order of an element divides the order of the group, so $|h|$ divides $|H|$.
We can use this view to argue as follows:
$$
\begin{align*}
h^2&\in\langle h\rangle\\
\Rightarrow \langle h^2\rangle&\leq\langle h\rangle&\text{by minimality of $\langle h^2\rangle$}\\
\Rightarrow |\langle h^2\rangle|&\leq|\langle h\rangle|\\
\Rightarrow |h^2|&\leq|h|
\end{align*}
$$
One can then show that if $|h|$ is odd then $\langle h^2\rangle=\langle h\rangle$, and so $|h^2|=|h|$. Otherwise, $\langle h^2\rangle\lneq\langle h\rangle$, and so $|h^2|\lneq|h|$.
