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This is a part in Dummit & Foote's abstract algebra on page 184.

In the middle of the example of groups of order $30$, the book lets $H=\langle a\rangle \times\langle b\rangle \cong \Bbb Z_5\times \Bbb Z_3$. It concludes that ${\rm Aut}(H)\cong{\rm Aut}(\Bbb Z_5)\times{\rm Aut}(\Bbb Z_3)$. The reason is explained in the book that "since these two subgroups are characteristic in $H$."

From my understanding, the explanation in the book implies that if $G=A\times B$ and $A$ and $B$ are characteristics in $G$, then ${\rm Aut}(G)={\rm Aut}(A)\times{\rm Aut}(B)$.

Is this true statement?

What I don't understand is how the characteristics of two subgroups matters.

I am a little baffled because there is no section in the book that talks about this. Can you please give me some insight?

jk001
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  • I am afraid I still have a question hanging. It seems that the statement holds without the restriction that $A$ and $B$ are characteristics in $G$. I don't know why characteristics matters as the book explains – jk001 Jul 07 '21 at 22:10
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    It is certainly not true that the statement holds for all groups $A$ and $B$. – Derek Holt Jul 07 '21 at 22:19
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    Sorry, with the restriction that the order of $A$ and $B$ is relatively prime – jk001 Jul 07 '21 at 22:25
  • I suggest you [edit] the question accordingly, @jk001. – Shaun Jul 07 '21 at 22:46
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    @jk001 If the subgroups have coprime order, then they are characteristic. – verret Jul 08 '21 at 03:43
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    In fact ${\rm Aut}(A \times B) = {\rm Aut}(A) \times {\rm Aut}(B)$ (that's equality, not just isomorphism) if and only if $A$ and $B$ are characteristic in $A \times B$, and the proof is straightforward. – Derek Holt Jul 08 '21 at 07:53
  • Thank you Derek. I will try to prove – jk001 Jul 08 '21 at 12:28

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