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Let $G$ and $H$ be finite groups. If $|G|$ and $|H|$ are coprime, then

$$\operatorname{Aut}(G \times H) \cong \operatorname{Aut}(G) \times \operatorname{Aut}(H)$$

holds. What about when $(|G|, |H|) > 1$? In this case we know that $\operatorname{Aut}(G) \times \operatorname{Aut}(H)$ is contained in $\operatorname{Aut}(G \times H)$, but the isomorphism above might not hold. For example $\operatorname{Aut}(C_2 \times C_2)$ has order $6$ but $\operatorname{Aut}(C_2) \times \operatorname{Aut}(C_2)$ is trivial.

Is the isomorphism possible at all when $(|G|, |H|) > 1$?

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I've merged the older question that this question was a duplicate of into this newer question because the newer title is much better and has a few more good answers. –  Alexander Gruber Jun 15 '13 at 20:06

4 Answers 4

up vote 14 down vote accepted

No, your condition is not enough.

For example, take two distinct finite nonabelian simple groups $G$ and $H$ such that neither one is a subgroup of the other (such pairs exist; the condition is in fact stronger than it needs to be, as witnessed by the theorem quoted in the end; you just need $G$ and $H$ to be distinct). By the odd-order theorem, both $|G|$ and $|H|$ are even, so $\gcd(|G|,|H|)\neq 1$.

Let $\iota_G$ and $\iota_H$ be the inclusions into the product, and $\pi_G$, $\pi_H$ the projections. Let $f\colon G\times H \to G\times H$ be an automorphism.

Then $\pi_H\circ f\circ\iota_G \colon G\to H$ is a homomorphism. Since $H$ does not contain a subgroup isomorphic to $G$ and $G$ is simple, the composition must be the trivial map. Therefore, $f(g,e)\in G\times\{e\}$ for all $g\in G$. Symmetrically, by looking at $\pi_G\circ f\circ \iota_H$, we conclude that $f(e,h)\in \{e\}\times H$ for all $h\in H$. Therefore, $f|_{G\times\{e\}} = \alpha\in \mathrm{Aut}(G)$, and $f|_{\{e\}\times H} = \beta\in \mathrm{Aut}(H)$. So every automorphism of $G\times H$ corresponds to an element of $\mathrm{Aut}(G)\times\mathrm{Aut}(H)$, and of course the restrictions completely determine $f$.

You may want to consider:

Bidwell, J.N.S., Curran, M.J., and McCaughan, D. Automorphisms of direct products of finite groups, Arch. Math. (Basel) 86 (2006) no. 6, 481-489

Bidwell, J.N.S. Automorphisms of direct products of finite groups II. Arch. Math. (Basel) 91 (2008) no. 2, 111-121.

An example of a theorem of the first one is:

Theorem. Let $G=H\times K$, where $H$ and $K$ have no common direct factor. Then $\mathrm{Aut}(G)\cong \mathcal{A}$, where $$\mathcal{A} = \left.\left\{\left(\begin{array}{cc}\alpha&\beta\\\gamma&\delta\end{array}\right)\;\right|\; \alpha\in\mathrm{Aut}(H), \delta\in\mathrm{Aut}(K), \beta\in\mathrm{Hom}(K,Z(H)), \gamma\in\mathrm{Hom}(H,Z(K))\right\}.$$ In particular, $$|\mathrm{Aut}(G)| = |\mathrm{Aut}(H)||\mathrm{Aut}(K)||\mathrm{Hom}(H,Z(K))||\mathrm{Hom}(K,Z(H))|.$$

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I’m no algebraist, but that strikes me as a very pretty theorem. –  Brian M. Scott Jun 13 '12 at 4:49
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@talmid: Multiplication of matrices, with the "product" of entries being composition. For the "sum", in the product $$\left(\begin{array}{cc}\alpha&\beta\\{\gamma}&\delta\end{array}\right) \left(\begin{array}{cc}a&b\\ c&d\end{array}\right)$$the (1,1) entry is $\alpha a + \beta c$. Note $\alpha a\in \mathrm{Aut}(H)$, and $\beta c\in \mathrm{Hom}(H,Z(H))$, so you can "add them" since $\beta c(g)$ is central. –  Arturo Magidin Jun 13 '12 at 5:15
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@talmid: If you check all the entries, you'll see they all make sense and lie where they need to lie. –  Arturo Magidin Jun 13 '12 at 5:21
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@talmid: Take $\alpha a+\beta c$. Given any $h,h'\in H$ we have that $\alpha a(h)\alpha a(h') \beta c(h)\beta c(h') = \alpha a(h)\beta c(h) \alpha a(h')\beta c(h')$, because $\beta c(h)\in Z(H)$. Thus, the map $x\mapsto (\alpha a(x))(\beta c(h))$ is a group homomorphism. So the "adding" is done by using the operation in the group. In each case, one of the "summands" lies in the center, which is what guarantees that we get a homomorphism even though the target group need not be abelian. –  Arturo Magidin Jun 13 '12 at 5:28
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@talmid: Additive notation so we don't confuse "multiplication" (really composition) of the functions, with pointwise multiplication of the functions (which in these cases works out as a homomorphism because at least one of the "summands" always takes value in the center). –  Arturo Magidin Jun 13 '12 at 5:32

It follows from the structure theory of automorphisms of direct products of finite groups that

$$|{\rm Aut}(G\times H)|=|{\rm Aut}(G)|\,|{\rm Aut}(H)|\,|\hom(G,Z(H))|\,|\hom(H,Z(G))|.$$

when $G$ and $H$ have no common direct factor. In particular, $(G,Z(H))=1=(Z(G),H)$ (along with the direct factor condition) is sufficient for ${\rm Aut}(G\times H)\cong {\rm Aut}(G)\times{\rm Aut}(H)$.

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If you look at the statement of the theorem, we need to assume that $G$ and $H$ have no common direct factor. For example, consider $G = H = S_3$. In this case $\operatorname{Aut}(G \times H)$ has order $72$ and $\operatorname{Aut}(G) \times \operatorname{Aut}(H)$ has order $36$. However, $Z(G) = Z(H) = 1$ so $|\hom(G,Z(H))| = |\hom(H,Z(G))| = 1$. Thus the formula does not hold in this case. –  Mikko Korhonen Jun 15 '13 at 8:12
    
@m.k. Yes you are correct. I was operating on incomplete memory of what I linked to. –  anon Jun 15 '13 at 8:14
    
Anyway, this seems like a pretty cool theorem. So for easy examples, just take $G$ and $H$ with no common direct factor such that $Z(G) = Z(H) = 1$ and $(|G|, |H|) > 1$. For example $G = S_3$ and $H = D_{10}$ works, in this case $G \times H$ has order $60$, perhaps this could be the smallest example. –  Mikko Korhonen Jun 15 '13 at 8:19
    
It would be nice to state explicitly whether this answer implies a yes or a no answer to the question. –  Marc van Leeuwen Jun 15 '13 at 8:58

Let $g\in \operatorname{Aut}(G)$, $h\in \operatorname{Aut}(H)$, $f\in \operatorname{Hom}(G,H)$. Then we have an automorphism $(x,y)\to (gx\cdot fy, hy)$. Therefore if $\operatorname{Hom}(G,H)\ne 1$ then $\operatorname{Aut}(G \times H) \ne\operatorname{Aut}(G) \times \operatorname{Aut}(H)$.

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Does this imply anything for the answer to the question? –  Marc van Leeuwen Jun 15 '13 at 8:59
    
@Marc van Leeuwen: Yes, since the number of triples $(f,g,h)$ is greater than $\operatorname{Aut}(G) \times \operatorname{Aut}(H)$ if $|\operatorname{Hom}(G,H)|> 1$. –  Boris Novikov Jun 15 '13 at 9:26
    
Two remarks: (1) you formula seems to assume $\def\Hom{\operatorname{Hom}}f\in\Hom(H,G)$ rather than $f\in\Hom(G,H)$; (2) unless the image of $f$ is central ($gx$ and $fy$ commute), I don't think this is a group (auto)morphism. –  Marc van Leeuwen Jun 15 '13 at 9:28
    
But $|\operatorname{Hom}(G,H)|>1$ does not follow from the hypothesis in the question. (I am assuming you meant to infer a positive answer to the question, but this is not the right answer.) –  Marc van Leeuwen Jun 15 '13 at 9:31
    
@Marc van Leeuwen: The question is: "Is the isomorphism possible...?" I give a partial answer: "It is impossible if..." Besides, I think that the correct formulation of the question should include $\def\Hom{\operatorname{Hom}}f\in\Hom(H,G)$. –  Boris Novikov Jun 15 '13 at 9:47

No, it does not follow.

Let $G,H$ be two finite non-cyclic simple groups with $|G|,|H|$ both even (this in always holds) but not multiple of one another (for instance one can take $|G|=60$ and $|H|=168$). Clearly $2\mid\gcd(|G|,|H|)$, but the subgroups $G,H$ of the product group $G\times H$ are characteristic subgroups (stable under all automorphims), so that every automorphism of $G\times H$ comes from the automorphisms it induces in the factors $G,H$ separately. To see that the factor subgroups are characteristic, it suffices to consider the image of one factor, say $G$, by an automorphism, and then projected to the other factor, $H$. The result is a subgroup of$~H$ isomorphic to a quotient of$~G$. But $H$ has no subgroup isomorphic to all of $G$ (consider the orders), and the only other quotient of $G$ is the trivial group (by simplicity of$~G$), so that is what the result is. This means that $G$ was stable under the automorphism. The same goes with $G$ and $H$ interchanged.

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My answer is virtually identical to this one by Arturo Magidin at an older identical question. Honestly, I did not look at it before writing this: I did not realise then that the link by anon was actually to a duplicate question. So paradoxically, I will now vote to close this one. –  Marc van Leeuwen Jun 15 '13 at 8:56

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