Isomorphisms between the groups $U(10), U(5)$ and $\mathbb{Z}/4\mathbb{Z}$

Solution 1:

General remark on group isomorphisms

Basically isomorphisms establish correspondance between the composition tables of two groups. If $(G,+)$ and $(H,+)$ are groups of the same size and we have an isomorphism $$ \varphi:G\overset{\sim}{\longrightarrow}H $$ then if we have elements $g_1,g_2\in G$ which composed gives $g_1+g_2=x\in G$ these are mapped via $\varphi$ to elements $h_1,h_2\in H$ that composed gives $h_1+h_2=y\in H$ in the way that $$ g_1+g_2=x\overset{\varphi}{\longmapsto}y=h_1+h_2 $$ or using the standard language of homomorphisms $$ \varphi(g_1+g_2)=\underbrace{\varphi(x)}_{y}=\underbrace{\varphi(g_1)}_{h_1}+\underbrace{\varphi(g_2)}_{h_2} $$ Now, since $\varphi$ is injective the image of $G$ in $H$ is just as detailed as $G$ itself, since any distinct elements of $G$ maps to distinct elements of $H$, so the entire composition table of $G$ is found as a copy in $H$. Furthermore an isomorphism is surjective so that all of $H$ is described via the composition tables of $G$.

The order of an isomorphically mapped element

In particular the isomorphism $\varphi$ has to carry compositional properties from $G$ to $H$. This means in particular that the order of $g\in G$ corresponds to the order of $\varphi(g)\in H$.

If $g$ has order $k$ in $G$ this means that $g^k=\underbrace{g+g+...+g}_{k\mbox{ times}}=0$ and that this $k$ is minimal in this respect so that $$ \{g,g^2,...,g^k\} $$ is a subset of $k$ distinct elements of $G$. Now since $$ \begin{align} \varphi(g^m)&=\varphi(g+g+...+g)\\ &=\varphi(g)+\varphi(g)+...+\varphi(g)\\ &=\varphi(g)^m \end{align} $$ and since an injective map maps $k$ distinct elements to $k$ distinct elements this means that $$ \{\varphi(g),\varphi(g)^2,...,\varphi(g)^k\} $$ is a subset of $H$ containing $k$ distinct elements. And in particular $$ g^k=0\overset{\varphi}{\longmapsto} 0=\varphi(g)^k $$ showing that $\varphi(g)$ has order $k$ in $H$. This shows why you should map a generator to a generator...

The composition tables in your examples

$$ \newcommand{\red}{\color{red}} \newcommand{\blue}{\color{blue}} \newcommand{\Z}{\mathbb{Z}} \begin{array}{c|c:c:c:c} \Z_4&\ 0\ &\ \red1\ &\ \blue2\ &\ \red3\ \\ \hline 0&0&\red1&\blue2&\red3\\ \hdashline \red1&\red1&\blue2&\red3&0\\ \hdashline \blue2&\blue2&\red3&0&\red1\\ \hdashline \red3&\red3&0&\red1&\blue2 \end{array} \quad \begin{array}{c|c:c:c:c} U(10)&\ 1\ &\ \red3\ &\ \blue9\ &\ \red7\ \\ \hline 1&1&\red3&\blue9&\red7\\ \hdashline \red3&\red3&\blue9&\red7&1\\ \hdashline \blue9&\blue9&\red7&1&\red3\\ \hdashline \red7&\red7&1&\red3&\blue9 \end{array} \quad \begin{array}{c|c:c:c:c} U(5)&\ 1\ &\ \red2\ &\ \blue4\ &\ \red3\ \\ \hline 1&1&\red2&\blue4&\red3\\ \hdashline \red2&\red2&\blue4&\red3&1\\ \hdashline \blue4&\blue4&\red3&1&\red2\\ \hdashline \red3&\red3&1&\red2&\blue4 \end{array} $$ In these tables I used colours to distinguish the order of the elements.