Isomorphisms in group theory

Today we’ll be taking off from before and talk about isomorphisms.

Let $G$ and $H$ be groups such that there is a map of the underlying sets $f:G\to H$.

Definition: The map $f:G\to H$ of the underlying sets is an isomorphism if $f$ is a bijection (i.e. one-to-one, i.e., the following two conditions are satisfied: 1) if $x=y$ in $G$ then $f(x)=f(y)$ in $H$, and 2) for every $y\in H$, there is some $x\in G$ such that $f(x)=y$. Alternatively, the map should have an inverse.) and $f$ respects the group structure, i.e., if $\bullet$ is the operation of $G$ and $\times$ is the operation of $H$, then $f(x\bullet y)=f(x)\times f(y)$. If there’s a map between groups $f:G\to H$ which is an isomorphism, we say that $G$ and $H$ are isomorphic, and we write $G\cong H$.

Remember that we usually drop the operation’s symbol, so $f(x\bullet y)=f(x)\times f(y)$ can be written as $f(xy)=f(x)f(y)$. What’s one condition for two groups to be isomorphic?

Proposition: If $G$ and $H$ are cyclic, and $|G|=|H|$, then $G\cong H$.

Proof: Since $G$ and $H$ are cyclic, write $G=\langle g\rangle$ and $H=\langle h\rangle$. Then define a map $f:G\to H$ which respects the multiplication as $g\to h$, so $f(g^i)=h^i$. The map $f$ obviously has an inverse, namely the map sending $h^i$ to $g^i$. We only need to check that $f(xy)=f(x)f(y)$. Since every $x,y$ is of the form $g^i, g^j$, we see that $f(xy)=f(g^{i+j})=h^{i+j}=h^ih^j=f(g^i)f^(g^j)$.

Another slightly more concrete isomorphism is the map $\mathbf{R}\to\mathbf{R}_{>0}$ defined by $x\mapsto e^x$. Now suppose $G$ is a general group. An isomorphism $G\to G$ is called an automorphism. For example, the map $f:G\to G$ defined by $x\mapsto gxg^{-1}$ is an automorphism. To see this it suffices to note that the inverse is $x\mapsto g^{-1}xg$, because $f$ obviously respects the group structure.

Let’s talk about something combinatorial. Let $S$ be a set. A permutation of $S$ is a bijection $S\to S$. For example, if $S$ is the set $\{1,\cdots,n\}$, the identity $S\to S$ defined by $x\mapsto x$ is a permutation, as is the map $x\mapsto x+1$ for $x>2$ and $n\mapsto 1$. Alright, so suppose we’ve got two permutations $f,g:S\to S$. Then the composition $f\circ g$ is a permutation. This is true because its inverse is just $g^{-1}\circ f^{-1}$. Also, if $f$ is a permutation, then so is its inverse $f^{-1}$. Let $\mathscr{P}(S)$ denote the set of permutations of $S$.

Lemma: $\mathscr{P}(S)$ is a group with the operation of composition.

Proof: We already stated that it’s closed under composition and inverses. The identity is, well, the identity function. Composition of functions is associative, so we’re done!

Now we’ll specialize to the case where $S=\{1,2,\cdots,n\}$. We claim that $\mathscr{P}(S)$ has $n!$ elements. We’ll prove this. If $k\in S$, then there are $n$ possibilities for which element of $S$ we can map $k$ to. Now let $k^\prime\in S$ such that $k^\prime\neq k$. Because the maps in consideration are bijections, $k^\prime$ can’t be mapped to the same element which $k$ is mapped to. So there are $n-1$ possibilities for which element of $S$ $k^\prime$ can be mapped to. Doing this for each element of $S$ gives us $n\times (n-1)\times\cdots\times 2\times 1$ possibilities for the permutations $S\to S$. Well, $n\times (n-1)\times\cdots\times 2\times 1=n!$, so we’re done!

If $S=\{1,2,\cdots,n\}$, we write $\mathscr{P}(S)$ as $S_n$ and call it the $n$th symmetric group. This actually isn’t abelian for $n>2$. Alright, so let’s take a look at $S_2$. This has $2!=2$ elements. Now this is a group of prime order, and so is cyclic. What about $\mathbf{Z}/2\mathbf{Z}$? This is also cyclic, and has order $2$. So we see, using the above result, that there’s an isomorphism $S_2\cong\mathbf{Z}/2\mathbf{Z}$. (In fact, because for higher values of $n$, $S_n$ isn’t of prime order (because $n!$ isn’t prime), it’s not cyclic for $n>2$.) There’s a very interesting (but not really helpful) result on general groups relating them to permutation groups:

Cayley’s theorem: Let $G$ be a group. Then $G$ is isomorphic to a subgroup of a permutation group.

Proof: Let $\phi:G\to \mathscr{P}(G)$ be the map which takes $g\in G$ to the following permutation $\sigma_g$ of $G$: $\sigma_g:a\mapsto ga$. We note that $\sigma_g$ is an isomorphism because it respects the multiplication and $\sigma_g$ has an inverse given by $a\mapsto g^{-1}a$.

Let $\mathrm{Im}(\phi)$ denote the image of $G$. This is a group because $G$ is a group. We want to show that $G$ is isomorphic to $\mathrm{Im}(\phi)$. We just need to check that the map $\phi:G\to \mathrm{Im}(\phi)$ is a bijection, because, as you can check, it respects the group structure (i.e. the multiplication). Recall that this means the following: 1) if $x=y$ in $G$ then $\phi(x)=\phi(y)$ in $\mathrm{Im}(\phi)$, and 2) for every $y\in \mathrm{Im}(\phi)$, there is some $x\in G$ such that $\phi(x)=y$.

First we check the first condition. Suppose $x=y$ in $G$. Then let $\sigma,\alpha$ denote the maps $\phi(x),\phi(y)$, respectively. We want to show that $\sigma=\alpha$. Well, what does this mean? It means that for any $g\in G$, $\sigma(g)=\alpha(g)$. But this is automatically true if $x=y$. Ok, now it is incredibly easy to see that for every $y\in \mathrm{Im}(\phi)$, there is some $x\in G$ such that $\phi(x)=y$, because by definition $\mathrm{Im}(\phi)$ satisfies this! So $G$ is isomorphic to $\mathrm{Im}(\phi)$, which is in turn a subgroup of the permutation group $\mathscr{P}(G)$.

Corollary: If $G$ is finite, then $G$ is isomorphic to a subgroup of $S_n$.

Proof: Trivial.

Here’s a cool isomorphism. Let $D_n$ denote the group of symmetries of a regular $n$-gon. For example, $D_3$ is simply the group of symmetries of an equilateral triangle. Drawing it out tells us that there are three rotation operations (namely: do nothing (the identity), rotate by $120^\circ$, and rotate by $240^\circ$) and three flipping operations (namely flipping around the line drawn through one of the vertices which bisects the opposite edge). In total, $|D_3|=6$. Wait a second! We also know that $|S_3|=1\times2\times3=6$. Is it possible that $D_3\cong S_3$? Yes! They are, indeed, isomorphic! This makes sense, obviously, because the elements of $S_3$ permute the vertices of the equilateral triangle, which is exactly what $D_3$ does!

We may spend more time on the symmetric group later, but we should get the notion of a homomorphism out of the way first, which generalizes the notion of an isomorphism. The map $\mathbf{Z}\to\mathbf{Z}/n\mathbf{Z}$ defined by taking $x\mapsto x\bmod n$ will be an example of a homomorphism (but note that it’s not an isomorphism!). That will be something for next time.

Have fun,
SKD