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If $H$, $K$ are characteristic subgroups of $G$, when is $\operatorname{Aut}(H\times K) = \operatorname{Aut}(H) \times \operatorname{Aut}(K)$ true?

What if $H$, $K$ are not characteristic subgroups?

In general, under what conditions is it true that $\operatorname{Aut}(H\times K) = \operatorname{Aut}(H) \times \operatorname{Aut}(K)$, for subgroups $H$, $K$ of $G$?

Librecoin
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FNH
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    charatistic? .... or characteristic? (invariant by all automorphisms) – Andrea Mori Jun 14 '13 at 23:43
  • it's characteristic , i will change it :) just missed the right spelling ! sorry for that – FNH Jun 14 '13 at 23:46
  • One particular case of your first question is: if $H$ is a characteristic subgroup of some group $G$, is $Aut(H\times H)\cong Aut(H)\times Aut(H)$, and the answer is no. For example, there exist groups $G$ whose center $H$ is cyclic of order $p$, and for them the isomorphism does not hold. It follows, of course, that the answer to your second question is also no. – Mariano Suárez-Álvarez Jun 14 '13 at 23:49
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    The question makes no sense, because you have not said what $G$ is. – Derek Holt Jun 15 '13 at 08:37

3 Answers3

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Regarding no restriction on two subgroups $H,K$ of $G$, but finiteness; if $(|H|,|K|)=1$ then $$Aut(H\times K)=Aut(H)\times Aut(K)$$ By setting $H=K=\mathbb Z_p$- so we have $(|H|,|K|)>1$- this property fails to be correct. Morover, it seems that if $G=H\times K$ then:

Let $f\in Aut(G)$. So $$f_H=f|_H\in Aut(H),~~f_K=f|_K\in Aut(K)$$ On the other hand, if $$f_H\in Aut(H),~~f_K\in Aut(K)$$ you may set $f=f_H\times f_K:HK\to HK$ with a rule (?) then $f\in Aut(G=HK)$ (?). Now assume that $$\psi:Aut(H)\times Aut(K)\to Aut(G),~~\psi(f_H,f_K)=f_H\times f_K\\\\ \phi: Aut(G)\to Aut(H)\times Aut(K),~~\psi(f)=(f_H, f_K)$$ Try to show that these are inverses to each other.

Mikasa
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  • can you add the proof of the property plz ? also , can you clarify why this doesn't work when $H=K=Z_p$ despite that $(|H|,|K|)=p$ ? – FNH Jun 15 '13 at 00:01
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    You have a typo: the first $p$ should be a $1$. – Mariano Suárez-Álvarez Jun 15 '13 at 00:06
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    @MarianoSuárez-Alvarez: Yes. Thanks it 'll be fixed. – Mikasa Jun 15 '13 at 00:13
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    Nice work, Babak! (+) – amWhy Jun 15 '13 at 00:24
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    Exactly the answer to the question. +1 – DonAntonio Jun 15 '13 at 00:24
  • @BabakS. , the rule must be - as i think - $f(hk)=f_H(h)f_K(k)$, right ? . you just define $G=HK$ or you try to say that the internal product of two characteristic gives the $HG$ where $HG$ means Hall Group :D – FNH Jun 15 '13 at 00:32
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    If $G=H \times K$, everything seems fine. If $G$ is larger, then $\phi$ need not be injective or surjective. – Jack Schmidt Jun 15 '13 at 00:45
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    @MathsLover: Yes the rule you pointed really works. I just see $G$ as $HK$. – Mikasa Jun 15 '13 at 08:04
  • @BabakS. , great , can you plz check my proof ? i added an answer , i don't know if this proof even makes sense for other " but i found upvotes ! " as i think it's one of the baddest proofs i have ever written ! – FNH Jun 15 '13 at 13:32
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I don't think the original question specified that $G = H \times K$. Without this specification, the answer is quite different (as Mariano indicated in the comments).

Consider $G = D_8 \times Q_8$. Then $H=Z(D_8) \times 1$ and $K=1 \times Z(Q_8)$ are both characteristic subgroups of $G$. The automorphism group of $G$ has order 3072, the automorphism group of $H \times K$ has order 6, and the image of the automorphism group of $G$ inside the automorphism group of $H \times K$ has order 1. In particular, the map $\phi$ from Babak's answer need not be injective or surjective if $G \neq H \times K$.

Let's look at homomorphisms of direct products.

Proposition: There is a correspondence between $\newcommand{\Hom}{\operatorname{Hom}}\Hom( H \times K, G )$ and $\Hom(H,G) \times \Hom(K,G)$ which is injective in general, and bijective if $G$ is abelian.

Proof: Suppose $f:H \times K \to G$ is a homomorphism. it turns out $f$ is also a direct product: $f((h,k)) = f((h,1) \cdot (1,k)) = f((h,1)) \cdot f((1,k))$. Define $f_H : H \to G : h \mapsto f((h,1))$. Since $(h_1,1)\cdot(h_2,1) = (h_1\cdot h_2,1)$ we get $f_H( h_1 \cdot h_2) = f_H(h_1) \cdot f_H(h_2)$ so $f_H$ is also a homomorphism. Similarly $f_K$ is a homomorphism.

Conversely, suppose we have homomorphisms $f_H : H \to G$ and $f_K : K \to G$ and $G$ is abelian, define $f: H \times K \to G : (h,k) \mapsto ( f_H(h), f_K(k) )$. Since $(h_1,k_1) \cdot (h_2,k_2) = ( h_1\cdot h_2, k_1 \cdot k_2)$ one has $f((h_1,k_1) \cdot(h_2,k_2)) = f((h_1,k_1)) \cdot f((h_2,k_2))$ and $f$ is also a homomorphism.

In summary: the correspondence takes $f$ to $(f_H,f_K)$ as in the first paragraph, and takes $(f_H,f_K)$ to $f$ as in the second paragraph as long as $G$ is abelian. $\square$

Proposition: There is a 1-1 correspondence between $\Hom(G,H \times K)$ and $\Hom(G,H) \times \Hom(G,K)$.

Proof: If $f:G \to H \times K$ is a homomorphism, define $f^H:G \to H:g\mapsto h$ and $f^K:G \to K: g \mapsto k$ where $f(g) = (h,k)$. It is not hard to show $f^H$ and $f^K$ are homomorphisms.

Conversely, if $f^H:G \to H$ and $f^K:G \to K$ are homomorphisms, then define $f:G \to H \times K : g \mapsto (f^H(g), f^K(g))$. It is not hard to show $f$ is a homomorphism.

In summary, the 1-1 correspondence takes $f$ to $(f^H,f^K)$ as in the first paragraph, and takes $(f^H,f^K)$ to $f$ as in the second paragraph. $\square$

Combining these we can handle the case where both the domain and range are direct products. Note that $(f^H)_K = (f_K)^H$ so I just write $f^H_K$.

Proposition: There is a correspondance between $\Hom(H \times K)$ and $\Hom(H,H) \times \Hom(H,K) \times \Hom(K,H) \times \Hom(K,K)$ that is injective in general, and bijective if $H,K$ are abelian.

Proof: $f \mapsto (f^H_H, f^K_H, f^H_K, f^K_K)$. The reverse direction holds if $H,K$ are abelian. $\square$

Now automorphisms are more complicated in general, but if $H$ and $K$ are reasonable independent (contain no isomorphic direct factors), then one has a nice expression given in Arturo's answer.

However, if $\Hom(H,K) = \Hom(K,H) = 0$, then we have diagonal matrices and automorphisms are easy to understand.

Community
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Jack Schmidt
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  • Another good choice of $G$ is $\newcommand{\SL}{\operatorname{SL}}G=\SL(2,5) \times \SL(2,7)$ with $H$ and $K$ the centers of the factors. Aut(G) = PGL(2,5) times PGL(2,7) is sent to the identity in Aut(H times K) = GL(2,2) – Jack Schmidt Jun 15 '13 at 00:59
  • @JackSchmidt , may you check my answer for my question ? i added an answer , you will find it at the end of the page . – FNH Jun 15 '13 at 14:14
  • There is a map $\hom(H\times K,G)\to\hom(H,G)\times\hom(K,G)$ given by $f\mapsto (f|_H,f|_K)$, where the vertical bar denotes restriction and we view $H$ and $K$ as subgroups of $H\times K$. This map is injective but not generally surjective. The reason is that $[H,K]=1$ inside $H\times K$, hence if $f$ is a map in $\hom(H\times K,G)$ then $[f(H),f(K)]=1$ inside $G$, however for $$(u,v)\in\hom(H,G)\times\hom(K,G)$$ there is no equivalent constraint of $[u(H),v(K)]=1$. But if (and only if) this constraint is satisfied, $(h,k)\mapsto u(h)v(k)$ is in fact a homomorphism $H\times K\to G$. – anon Jun 16 '13 at 01:20
  • In category theory we know that $\hom(\coprod X,Y)\cong\prod\hom(X,Y)$, where $\coprod$ and $\prod$ are respectively coproduct and product in an arbitrary category, so the first proposition here would at face value seem correct since direct sum and direct product are isomorphic when there are a finite number of summands/factors involved. The issue for this situation is that $\times$ is actually not a coproduct in $\sf Grp$; direct sum $\oplus$ is the coproduct in $\sf AbGrp$ and the free product $*$ is the coproduct in $\sf Grp$. – anon Jun 16 '13 at 01:22
  • @anon fixed. the application only needed the injectivity to represent a homomorphism as a matrix. Not all matrices represent homomorphisms. – Jack Schmidt Jun 16 '13 at 04:23
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it's the first time i try to answer my question ! i don't know if this will make the question closed or what will happen ,but i have a trial to prove this statement and want to share it with you ,and i hope there is no logical mistakes in it .

to prove $Aut(H \times K) = Aut(H) \times Aut(K) $

1- i prove that $Aut(H \times K) \subseteq Aut(H) \times Aut(K) $

let $M \in Aut(H \times K)$ then $M=(M_1,M_2)$

$M_1(H) \times M_2(K) = (M_1,M_2)(H\times K)=M(H\times K)=H\times K $

so , $M=(M_1,M_2) \in Aut(H)\times Aut(K)$

so, $Aut(H \times K) \subset Aut(H) \times Aut(K) $

2- i prove that $Aut(H) \times Aut(K) \subseteq Aut(H \times K) $

let $M\in Aut(H)\times Aut(K) so M=(M_1,M_2) where M_1 \in Aut(H)$ and $M_2 \in Aut(K) $

$M(H\times K)=(M_1,M_2)(H\times K)=(M_1(H)\times M_2(K))=(H,K)$

so$M\in Aut(H\times K)$

so , $Aut(H) \times Aut(K) \subseteq Aut(H \times K) $

using 1,2 ,we conclude that ,$Aut(H \times K) = Aut(H) \times Aut(K) $

i think that this is the Baddest proof i have ever written ! i don't know if it even makes sense !

FNH
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    However, I personally would prefer you asked Jack, it seems to me right. Though, there is a differences between the notation I employ and the ones, you took. ;) – Mikasa Jun 15 '13 at 14:11
  • @BabakS. , thanx :) i will ask him :) thank you very much :) – FNH Jun 15 '13 at 14:13
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    If $M \in \operatorname{Aut}(H \times K)$ then $M$ is like a $2 \times 2$ matrix. $$M(h,k) = M(h,1) \cdot M(1,k) = ( M_{11}(h), M_{12}(h) ) \cdot ( M_{21}(k), M_{22}(k) ) = ( M_{11}(h) \cdot M_{21}(k) , M_{21}(h) \cdot M_{22}( k) )$$ In your answer, you seem to leave out $M_{12}: H \to K$ and $M_{21}:K \to H$. If $\gcd(|H|,|K|) = 1$, then this is OK, since $M_{12}(h)=1$ and $M_{21}(k)=1$, but otherwise it is probably not right. – Jack Schmidt Jun 15 '13 at 14:29
  • @JackSchmidt , in reality , i'm not familiar with making connection between groups and matrices and how they interchange their action on other mathematical structures so i don't understand why should $M$ and the rest of $M$'s act in this way nor why the steps go in this particular way , and of course , i can't ask you to clarify all those things up as it seems to be big deal and need so much effort , so i ask a reference or a link which i can look it up to understand what is going on . – FNH Jun 15 '13 at 20:38
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    @MathsLover: matrix stuff added to my answer. Just imagine $H$ and $K$ are 1-dimensional vector spaces with bases ${e_1}$ and ${e_2}$ and think about what a linear transformation looks like (a matrix :-) – Jack Schmidt Jun 15 '13 at 20:56
  • @JackSchmidt , i think i have to study more on linear algebra ! in reality , not so familiar with transformation of vector spaces ! i will learn more about those things first . i think this will enable me to look at groups from new perspective , right ? – FNH Jun 15 '13 at 23:25