Why is $U(10)\not\approx U(12)$? I'm trying to understand this example on isomorphism but failing to do so. I know that if two groups have a differing number of elements of each order they are not isomorphic. I assume this is what they are saying. Particularly, how is it that $\phi(3)\phi(3)$ and $\phi(1)\phi(1)$ both equal 1?

 A: As written explicitly, in $U(12)$ we have $g^2=1$ for every element $g$. Hence, $\phi(3)^2=\phi(1)^2=1$.
However, what you wrote about differing numbers of elements of the same order is indeed a simpler way to think of it. In $U(10)$ there are elements of order $4$, and in $U(12)$ there are no such elements. Consequently, the two groups cannot be isomorphic to one another.
A: $$\begin{align*}
U(10)&=\{x\in\mathbb{Z}/10\mathbb{Z}:x\text{ is a unit}\}=\{1+10\mathbb{Z},\;3+10\mathbb{Z},\;7+10\mathbb{Z},\;9+10\mathbb{Z}\}\\
U(12)&=\{x\in\mathbb{Z}/12\mathbb{Z}:x\text{ is a unit}\}=\{1+12\mathbb{Z},\;5+12\mathbb{Z},\;7+12\mathbb{Z},\;11+12\mathbb{Z}\}\\
\end{align*}$$
Both $U(10)$ and $U(12)$ are groups with $4$ elements.
(For convenience, we often drop the coset notation, but remember that it's still there implicitly.)
As is explained in the excerpt, we have $x^2=1$ for every $x\in U(12)$. That is,
$$\left.\begin{align*}
(1+12\mathbb{Z})^2&=1+12\mathbb{Z}\\
(5+12\mathbb{Z})^2&=25+12\mathbb{Z}\\
(7+12\mathbb{Z})^2&=49+12\mathbb{Z}\\
(11+12\mathbb{Z})^2&=121+12\mathbb{Z}
\end{align*}\right\}\text{ all equal }1+12\mathbb{Z}$$
(In other words, $25\equiv 1\bmod 12$, and $49\equiv 1\bmod 12$, and $121\equiv 1\bmod 12$.)
We suppose, for the sake of contradiction, that $\phi:U(10)\to U(12)$ is a function that is a group isomorphism. In particular, for any input $a\in U(10)$, the output $\phi(a)$ is an element of $U(12)$, and (as observed) therefore must have the property that $\phi(a)^2=1$.
But since
$$\begin{align*}
\phi(9+10\mathbb{Z})=\phi\biggl((3+10\mathbb{Z})^2\biggr)=\phi(3+10\mathbb{Z})^2=1+12\mathbb{Z}\\
\phi(1+10\mathbb{Z})=\phi\biggl((1+10\mathbb{Z})^2\biggr)=\phi(1+10\mathbb{Z})^2=1+12\mathbb{Z}\\
\end{align*}$$
(No matter what the value of $\phi(3+10\mathbb{Z})$ is! all we need to know is that it's an element of $U(12)$!) we see that $\phi$ is not injective, and therefore cannot be a group isomorphism as was assumed.
