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We were told in class that the automorphism of $\mathbb{Z_3} \times \mathbb{Z}_5$ is congruent to $\mathbb{Z}_2 \times \mathbb{Z}_4$.

I know that $Aut(G \times H) \cong Aut(G) \times Aut(H)$.

However I am confused at the idea what isomorphisms do the elements of $\mathbb{Z}_2 \times \mathbb{Z}_4$ represent?

What does $(1,0)$ represent? -is it the trivial isomorphism ?

What does $(1,1)$ represent ?

Can someone help me out with explicit representation of the elements and the automorphisms

Antimony
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    (i) It is false that ${\rm Aut}(G \times H) \cong {\rm Aut}(G) \times {\rm Aut}(H)$ in general: it works when the orders of $G$ and $H$ are relatively prime, but it's always false if $G = H$ and $G$ is nontrivial (consider the swap mapping $(g,g') \mapsto (g',g)$. (ii) The use of the notation $\mathbf Z_2$ and $\mathbf Z_4$ from your class is really terrible: you should instead refer to invertible elements (units) mod 3 and 5: $\mathbf Z_3^\times$ and $\mathbf Z_5^\times$, which are cyclic of order 2 and 4 but are not literally $\mathbf Z_2$ or $\mathbf Z_4$ (just isomorphic to these). – KCd Apr 30 '22 at 03:11
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    In short, do you understand why ${\rm Aut}(\mathbf Z_m)$ can be thought of as $\mathbf Z_m^\times$ (units mod $m$)? – KCd Apr 30 '22 at 03:12
  • Yes we map $1_m$ must be mapped to $k$ where $(k,m)=1$.So we get isomorphism between $Z_m^*$ and $Aut(Z_m)$ – Antimony Apr 30 '22 at 03:24
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    Then that basically solves the problem if you don't use the awful notation from class: write $\mathbf Z_5^\times$, not $\mathbf Z_4$. They're isomorphic, but not in an intrinsic way: $1 \bmod 4$ does not correspond to just one possible element of $\mathbf Z_5^\times$ (there are two possible generators). – KCd Apr 30 '22 at 03:50

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First, as KCd wrote in the comments, it's not generally true that $\operatorname{Aut}(G \times H) \cong \operatorname{Aut}(G) \times \operatorname{Aut}(H)$, even for cyclic groups $G, H$; we only necessarily have the apparent containment $\operatorname{Aut}(G \times H) \supseteq \operatorname{Aut}(G) \times \operatorname{Aut}(H)$.

Example ($G = H = \Bbb Z_2$). We have $\operatorname{Aut}(\Bbb Z_2 \times \Bbb Z_2) \cong S_3$ (here $S_3$ acts by permutations of the $3$ nonidentity elements), whereas $\operatorname{Aut}(\Bbb Z_2) \times \operatorname{Aut}(\Bbb Z_2)$ is trivial. (More generally, $\operatorname{Aut}(\Bbb Z_p^k) \cong \operatorname{GL}(k, \Bbb Z_p)$, whereas $\operatorname{Aut}(\Bbb Z_p)^k$ consists just of the diagonal elements of $\operatorname{GL}(k, \Bbb Z_p)$.)

It is true, however, that $\operatorname{Aut}(G \times H) \cong \operatorname{Aut}(G) \times \operatorname{Aut}(H)$ in the special situation that $|G|, |H|$ are coprime, which in particular applies to our case, $G = \Bbb Z_3$, $H = \Bbb Z_5$.

Characterization of the (group) automorphisms of $\Bbb Z_n$. Since the groups $\Bbb Z_n$ are cyclic, namely, generated by $\bar 1$, any automorphism $\phi \in \operatorname{Aut}(\Bbb Z_n)$ is characterized by the value $$\bar a := \phi(\bar 1) .$$ Put another way, (in terms of the usual ring multiplication of $\Bbb Z_n$) $\phi$ is just multiplication by $\bar a = \phi(\bar 1)$. One can check that this multiplication map is an automorphism iff (in fact, a bijection iff) $a$ and $n$ are coprime, so we can identify $\operatorname{Aut}(\Bbb Z_n)$ with the group $\Bbb Z_n^\times$ of multiplicative units of the ring $\Bbb Z_n$, that is, the set of elements of $\Bbb Z_n$ with a multiplicative inverse.

It thus follows from the comment after the example that $$\operatorname{Aut}(\Bbb Z_3 \times \Bbb Z_5) \cong \operatorname{Aut}(\Bbb Z_3) \times \operatorname{Aut}(\Bbb Z_5) \cong \Bbb Z_3^\times \times \Bbb Z_5^\times .$$

Now, it's true for prime $p$ that $\operatorname{Aut}(\Bbb Z_p) \cong \Bbb Z_p^\times \cong \Bbb Z_{p - 1}$, and so $\operatorname{Aut}(\Bbb Z_3 \times \Bbb Z_5) \cong \Bbb Z_2 \times \Bbb Z_4$, but the latter notation can create confusion, and in most (all?) situations using the multiplication group notation $\Bbb Z_n^\times$ will be clearer. In any case under that isomorphism, e.g.,

  • $(\bar 0, \bar 0)$ is the trivial automorphism,
  • $(\bar 1, \bar 0)$ is the involution (i.e., automorphism whose square is the identity), $(\bar x, \bar y) \mapsto (-\bar x, \bar y)$, and
  • $(\bar 1, \bar 1)$ can be identified with $(\bar x, \bar y) \mapsto (-\bar x, \bar 2 \cdot \bar y)$.

(In fact, $\Bbb Z_5^\times \cong \Bbb Z_4$ itself has two automorphisms, so we can identify $\bar 1 \in \Bbb Z_4$ with either automorphism of $\Bbb Z_5$ of order $4$, namely $\bar y \mapsto \pm \bar 2 \cdot \bar y$.)

Travis Willse
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Automorphisms of $\mathbb{Z}_n$ are determined by what $1$ is mapped to. We can map $1$ to $m \in \mathbb{Z}_n$ where $(m,n)=1$. Hence in $\mathbb{Z}_3$ and $\mathbb{Z}_5$ there are $2$ and $4$ choices respectively. Let $\rho_i \quad( i=1,2)$ and $\tau_j \quad (j\in \{1,2,3,4\})$ be automorphisms of $\mathbb{Z}_3$ and $\mathbb{Z}_5$ so that $\rho_i: 1\mapsto i$ and $\tau_j: 1\mapsto j$. Then $\{\rho_i\}$ and $\{\tau_j\}$ are cyclic groups of orders $2$ and $4$ respectively.

In general for $\mathbb{Z}_n$ automorphism group is of order $\phi(n)$, since $1$ is mapped to a generator of $\mathbb{Z}_n$.

In other words $Aut(\mathbb{Z}_n)\cong \mathbb{Z}^\times_n$; Hence in your question (1,1) represents $(\rho_1, \tau_1)$.

nafise modaresi
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