Finding all isomorphisms between two groups

Question

I understand what an isomorphism is. However, what does it mean to find all the isomorphisms between two groups?

For instance, if I am asked to find all the isomorphisms from $\mathbb{Z_{15}}$ to $\mathbb{Z_{15}}$, what would I have to do?

Answer 1

Here is an example of finding all homomorphisms between two groups. Let $G$ be an arbitrary group and consider all homomorphisms $\mathbb{Z} \to G$. We want to classify the homomorphism in such a way that (i) we can (more or less) count the number of homomorphism $\mathbb{Z}\to G$ and (ii) describe what each of this homomorphism does to elements of $\mathbb{Z}$ (i.e., where it sends them in $G$.

In this particular case, we note that the only thing that really matters in constructing a homomorphism $\mathbb{Z}\to G$ is where we send the integer $1$. I won't prove this here, it is something you should do yourself. But the reason why is essentially that $\mathbb{Z}$ is generated by $1$, so if we send $1$ to, say, $g\in G$, then we are forced to send the integer $n$ to $g^n$.

This tells us that the homomorphism $\mathbb{Z}\to G$ are in bijection with $G$ itself. So the cardinality of $\mathbb{Z}\to G$ is the cardinality of $G$. We also know what each of these does to the elements of $\mathbb{Z}$. It sends $1$ to an arbitrary element of $G$ and sends $n$ to the $n$-th power of said element.

We can restrict our attention to isomorphisms. A necessary condition on an isomorphism is that $1$ is sent to a generator of $G$ (though this is not sufficient). To make things simple, consider $G:\equiv \mathbb{Z}$ and lets count the isomorphisms $\mathbb{Z}\to\mathbb{Z}$. We know that we can construct a homomorphism by sending $1$ to an arbitrary integer. But to construct an isomorphism, we know $1$ must be sent to a generator (this is a fact you would need to prove). But there are only two generators of $\mathbb{Z}$, namely $1$ and $-1$. So there are two candidates for possible isorphisms $\mathbb{Z}\to\mathbb{Z}$: the first is the identity, which sends $1$ to $1$; the second inverts every element, since it sends $1$ to $-1$. It is easy to verify these are both isomorphisms. Thus there are exactly two isomoprhisms $\mathbb{Z}\to \mathbb{Z}$.

To count the isomorphism $\mathbb{Z}\to\mathbb{Z}$, I used a fact peculiar to $\mathbb{Z}$. But I hope the idea of counting isomorphisms and homomorphisms is now clear enough that you can take a crack at counting the isos $\mathbb{Z}_{15}\to\mathbb{Z}_{15}$.

Answer 2

There's $\varphi (15)=8$ of them, in this case, since generators are sent to generators.

They actually form a group, with composition as the operation, $\rm{Aut}(Z_{15}),$ or $\rm{Aut}(G),$ more generally, because an endomorphism that is an isomorphism is an automorphism.

In general, the number of isomorphisms between two groups will be the order of the automorphism group of either of them.

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To determine the group in the case of $\Bbb Z_{15}$ (even though you may consider this overkill), note that the automorphism group of a cyclic group $C_n$ is, largely because of the fact about generators, isomorphic to the group of units, or the multiplicative group $\pmod n.$ So in this case $\Bbb Z_{15}^×.$

Further, since $\Bbb Z_{15}\cong \Bbb Z_5×\Bbb Z_3$ and because of some category theory the group of units functor respects products, we get $$\Bbb Z_5^××\Bbb Z_3^×\cong \Bbb Z_4×\Bbb Z_2.$$