# How is $\operatorname{Aut}(\mathbb{Z}_4) = \mathbb{Z}^\times_4$?

If I understood it correctly, out textbook says that $\operatorname{Aut}(\mathbb{Z}_4)=\mathbb{Z}^\times_4$. I'm a little confused with that, because I thought automorphisms were "functions". How are elements of $\mathbb{Z}^\times_4$ "functions"?

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Automorphisms form a group, and this is just giving the specific group $\mathrm{Aut}(\Bbb Z/4)$ is isomorphic to. – Henry T. Horton Mar 19 '13 at 20:48
Given any $a\in{\bf Z}/n{\bf Z}$ (as a ring), the map $\varphi_a:x\mapsto ax$ is a well-defined automorphism of ${\bf Z}/n{\bf Z}$ (as an additive group); $a\leftrightarrow\varphi_a$ is an isomorphism $({\bf Z}/n{\bf Z})^\times\cong{\rm Aut}({\bf Z}/n{\bf Z})$. – anon Mar 19 '13 at 20:50
As Henry said, $\text{Aut}(\mathbb{Z}_4)$, as a group of group homomorphisms under function composition, is isomorphic to $\mathbb{Z}_4^\times$. The symbol $\cong$ would be more appropriate than $=$. – Alex Petzke Mar 19 '13 at 20:51

The notation $Z_4^x$ presumably means the set of elements $a \in Z_4$ such that there is $b$ in $Z_4$ with $ab = 1$. (But let me write $Z_4^{\times}$ instead: it looks a little nicer.) Here we are using the multiplicative structure of $Z_4$...so in fact I'd prefer to write it as $\mathbb{Z}/4\mathbb{Z}$ and explicitly view it as a ring.

Now for $a \in \mathbb{Z}/4\mathbb{Z}^{\times}$ multiplication by $a$ gives a function from $\mathbb{Z}/4\mathbb{Z}$ to $\mathbb{Z}/4\mathbb{Z}$: $g \mapsto ag$. That this is a group homomorphism is just distributivity: $a(g+g') = ag + ag'$. Moreover, this homomorphism has an inverse because $a$ does: there is $b \in \mathbb{Z}/4\mathbb{Z}^{\times}$ such that $ab = ba = 1$, and thus multiplication by $b$ is the inverse homomorphism to multiplication by $a$.

In fact, for any ring $R$ and any $a \in R^{\times}$ -- i.e., such that there is $b \in R$ with $ab = ba = 1$ -- the map $g \mapsto ag$ is an automorphism of the additive group $(R,+)$. (The map $g \mapsto ga$ is also an automorphism of the additive group: the same one if and only if $a$ lies in the center of $R$, so e.g. if $R$ is commutative.)

In general multiplication by a unit in a ring $R$ gives only a subgroup of the full automorphism group of the additive group $(R,+)$. But in this case, as your textbook says, you are getting the full automorphism group. In fact, for any positive integer $n$, we have

$\operatorname{Aut} (\mathbb{Z}/n\mathbb{Z},+) \cong (\mathbb{Z}/n \mathbb{Z})^{\times}$.

It is a nice exercise to prove this, and perhaps I should leave it to you. One way of getting started is being given an arbitrary automorphism $\varphi$ of $(\mathbb{Z}/n\mathbb{Z},+)$ and thinking about the element $\varphi(1)$...

Having done this you might try to work out the automorphism groups of $(\mathbb{Z},+)$ and $(\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z},+)$.

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For which (commutative) rings we have $Aut(R) = R^{\times}$? – lhf Mar 19 '13 at 21:43