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Let $D_{8}$ be group dihedral of order 8 and $C_{2}$ be cyclic group of order $2$. Then determine the number all automorphisms of $C_{2}\times D_{8}$. Can you determine automorphisms group of $C_{2}\times D_{8}$?

In general case we know that $\lvert\operatorname{Aut}(H\times K)\rvert\geqslant\lvert \operatorname{Aut}(H)\rvert\times\lvert \operatorname{Aut}(K)\rvert$ for finite groups $H$ and $K$. Also we know that if $(\lvert H\rvert, \lvert K\rvert)=1$, then $\lvert\operatorname{Aut}(H\times K)\rvert=\lvert \operatorname{Aut}(H)\rvert\times\lvert \operatorname{Aut}(K)\rvert$. If $(\lvert H\rvert, \lvert K\rvert)\neq 1$, then can we say that $\lvert\operatorname{Aut}(H\times K)\rvert>\lvert \operatorname{Aut}(H)\rvert\times\lvert \operatorname{Aut}(K)\rvert$?

Thank you

Davide Giraudo
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maryam
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  • please guide me. What do i must do? – maryam Jan 04 '13 at 13:58
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    If someone has answered a question of yours to your satisfaction, you can "accept" the answer by clicking on the check mark next to the answer. If you want to know more about this, I'm sure it is discussed in the faq, and there is a link to the faq at the top of this very page. – Gerry Myerson Jan 04 '13 at 14:07
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    Maryam never explains where the problems she asks come from, and does not usually respond to replies and queries. This seems to me to be a difficult question. A computer calculation shows that the order of the automorphism group is 64. – Derek Holt Jan 04 '13 at 14:13
  • Khayle mamnoon Dr Sorouh – maryam Jan 04 '13 at 14:19
  • @maryam: I wanted to note some important points about living in M.S.E. You can go back and see your previous questions to find out what I am telling you. Many Masters and Professors who you and me CANNOT meet them physically, have considered your problems kindly. Some of them are the editors of best journals in the world, nevertheless; they have tried to help you, guide you in a way they preferred. So, they expect you to write the problem clearly and show your attempts in solving the problem and noting the source you got the problem. – Mikasa Jan 04 '13 at 18:49
  • @maryam:They expect you to be involved with them during solving. I have seen that being silence is absolutely wrong here. It is my honor someone like Derek.H. left comments for my small problem, as he did for you very elegantly; so for surviving here do what I told you by words. Welcome here, maryam. And I amnot a Dr. we are just friens here. ;-) – Mikasa Jan 04 '13 at 18:49
  • I'm getting lots of hits on google for "Khayle mamnoon" but struggling to find a translation. Does it mean thank you very much? – Derek Holt Jan 04 '13 at 21:10
  • @DerekHolt: yes it does, prof. – Mikasa Jan 07 '13 at 06:36

1 Answers1

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This is not a complete answer.

Now $C_2 = \langle \alpha \rangle$ and $D_8 = \langle x , y\rangle$ where $\alpha$ and $x$ have order $2$ and $y$ has order $4$. Let $G = C_2 \times D_8$. For clarity, I'll denote $(\alpha, 1)$ by $\alpha$, $(1,x)$ by $x$ and $(1,y)$ by $y$ here. Because $G$ is generated by $\alpha, x$ and $y$, any automorphism $\phi$ of $G$ is completely determined by $\phi(\alpha)$, $\phi(x)$ and $\phi(y)$.

It is not difficult to show that $y^2$ is fixed by every automorphism of $G$. Therefore since $\alpha$ is central of order $2$, there are $2$ possible choices for $\phi(\alpha)$. The element $x$ is noncentral of order $2$, so there are $8$ possible choices for $\phi(x)$. The element $y$ has order $4$ so there are $4$ possible choices for $\phi(y)$.

Thus the group $G$ has at most $2 \times 8 \times 4 = 64$ automorphisms. According to GAP, there are exactly $64$ automorphisms, so it turns out these are all the automorphisms of $G$. I haven't figured out any nice way to show that each of the possible choices for $\phi(\alpha), \phi(x)$ and $\phi(y)$ determine an automorphism.

Mikko Korhonen
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    For a "nice way" to show your last line, write down a presentation for the group and verify that the maps still satisfy the presentation. Indeed, as the inner automorphisms are normal in the automorphism group you only need to check a single representative for each coset (there are $16$ such representatives, I believe). Then, you only need to check the cosets corresponding to generators of the outer automorphism group. – user1729 Jan 04 '13 at 16:17
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    (I should say - a presentation for this group is $\langle x, y, \alpha; \alpha^2, x^2, y^4, xyx=y^{-1}, \alpha x=x\alpha, \alpha y=y\alpha \rangle$. And that the above idea shows that the map is a homomorphism. If they are generators - which if you are careful they will be - you have a surjective homomorphism, which is an isomorphism because your group is finite. For infinite groups you need to also show that the map is injective. This is related to the ideas of a Hopfian group.) – user1729 Jan 04 '13 at 16:22
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    You can check that the following images of $(\alpha,x,y)$ all define automorphisms, and it is not hard to see that these are sufficient to generate all 64 automorphisms. $(\alpha y^2,x,y)$, $(\alpha,xy^2,y)$, $(\alpha,x,y^3)$, $(\alpha,x\alpha,y)$, $(\alpha,x,y\alpha)$, $(\alpha,xy,y)$. I wonder whether maryam will accept your answer? – Derek Holt Jan 04 '13 at 16:59
  • Khayle mamnoon professor Holt for your comment. – maryam Jan 06 '13 at 08:09