I'm working my way through Mac Lane and Birkhoff Algebra and I have a question about one of their exercises. We are asked to show that certain automorphism groups are isomorphic (as groups) to a given group. For example, one part is to show that $\operatorname{Aut} (\mathbb{Z}_6) \cong \mathbb{Z}_2$. I'm fairly confident that I can write out all of the automorphisms and show that they are the only ones and show that they form a cyclic group of order (which we know from the reading is isomorphic to $\mathbb{Z}_2$). But, I'd like to argue this in a different way, that might save some time for the other parts of the problem. I'm curious if my solution is valid, or if I am missing something(s). Here it goes.

From a prior problem (coincidentally that I asked about a few days ago), we know that if $\phi: G \rightarrow H$ is an morphism between groups, then the image of $\phi$ forms a subgroup, $\operatorname{Im}(\phi) \subseteq H$. Next, since subgroups of cyclic groups are cyclic, we know that $\operatorname{Im}(\phi) = \langle a \rangle$, for some $a\in H$. Finally, if we are talking about automorphisms then, in particular, $\phi$ is an epimorphism. So, $\langle a \rangle =\operatorname{Im}(\phi) = H$. (Does this imply that generators of $G$ must be mapped to generator(s) of $H$?)

Thus, this problem reduces to finding the number of distinct generators of each group in question. For example, $\mathbb{Z}_6$ has two generators, $1$ and $5$. Hence, there are two distinct automorphisms, implying that the order of $\operatorname{Aut} (\mathbb{Z}_6) $ is two. The first is the identity automorphism, call it $\phi_{1}$ and the second, call it $\phi_2$ sends: $$ \begin{align} 1 &\mapsto 5\\ 2 &\mapsto 2\\ 3 &\mapsto 3\\ 4 &\mapsto 4\\ 5 &\mapsto 1\\ \end{align} $$ From this, $\phi_2 \circ \phi_2 = \phi_1$. So, $\operatorname{Aut} (\mathbb{Z}_6) = \left\{ \phi_1, \phi_2 \, | \, \phi_2^2 = \phi_1\right\} \cong \mathbb{Z}_2$. (Could we not get another automrphism by taking $\phi_2$ and instead of sending $2 \mapsto 2$ and $4 \mapsto 4$, send $2 \mapsto 4$ and $4 \mapsto 2$?)

As you can see, this solution/these ideas are not fully baked, so any assistance would be greatly appreciated. Thanks in advance.


1 Answer 1


Question 1: Yes: if $\varphi\colon G\to H$ is a surjective group homomorphism${}^*$, and if $X\subseteq G$ generates $G$, then $\varphi(X)$ generates $H$. (More generally, for any group homomorphism, $\varphi(X)$ generates $\mathrm{Im}(\varphi)$). In the special case where $G$ is cyclic, then the image of a generator of $G$ must be a generator of $H$ (or of the image).

Question 2: Unfortunately, your map is not an automorphism, because it is not a homomorphism. Note that if $\phi(1)=a$, then $\phi(2)$ is forced: we must have $$\phi(2) = \phi(1+1) = \phi(1)+\phi(1) = a+a = 2a.$$ Thus, if $\phi(1) = 5$, then you must hvae $\phi(2) = 5+5 = 4$ (in $\mathbb{Z}_6$); similarly, $\phi(3) = 5+5+5=3$ (in $\mathbb{Z}_6$), and $\phi(4) = 2$. That is, in fact the “other” automorphism is the one you ask about, not the one you give.

In general, if you know what happens to a generating set, then this completely determines what happens to everyone else: because every other element can be written as a product of elements of $X$ and their inverses (or sums and difference in additive notation), and so the group homomorphism property tells you what that elements must be mapped to (the corresponding product of images and their differences).


${}^*$ I don’t like to use “epimorphism” as a synonym for “surjective” because epimorphism is a right cancellable morphism; in the category of all groups (and in natural categories of groups) all epimorphisms are surjective, but there are both classes of groups where they are not, and there are many natural categories where they are not (such as the category of monoids, semigroups, rings, rings with identity, Hausdorff topological spaces, and more). In fact, a good chunk of my doctoral dissertation was about nonsurjective epimorphisms in varieties of groups.

  • $\begingroup$ Ah yes. I see now. Thank you! $\endgroup$
    – raynea
    Jun 7, 2020 at 22:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .