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Regarding to this answer, I am looking for the presentation of $\mathbb Z_n\rtimes _{\phi}Q_8$, where
$$Q_8=\langle a,b\mid a^4=1, a^2=b^2, ba=a^3b\rangle=\{1,a,a^2,a^3,b,ab,a^2b,a^3b\}, ~~\mathbb Z_n=\langle c\rangle.$$

Here, for simplicity for myself, I set $n=3$, so : $$\mathbb Z_3\rtimes _{\phi}Q_8 \;=\; \langle a,b,c \mid a^4=1, a^2=b^2, ba=a^3b, c^3=1, yx=\phi_y(x)y\rangle$$ in which $y\in Q_8,~~x\in\mathbb Z_3$ and $\phi: Q_8\to Aut(\mathbb Z_3)$. The problem is really to define this homomorphism $\phi$ appropriately. Clearly, $\phi_y(1)=1$ and I have $$\phi_{a^i}(x)=?,~~\phi_{a^jb}(x)=?,~~\phi_b(x)=?, 1\leq i\leq3,~~1\le j\le 3$$ I thinking to myself to define $\phi_{a^i}(x)=id_{\mathbb Z_3}$ just to get rid of one part. Am I on a right way? Any suggestions?

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Mikasa
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2 Answers2

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This just comes down to working out who $Q_8$ maps to the cyclic group of order two (which is isomorphic to $\operatorname{Aut}(\mathbb{Z}_3)$). If you want the action to be non-trivial, you want the map to be onto (although this is not generally true, you just need the image to be non-trivial).

So, write $Q_8=\langle i,j,k,x; i^2=j^2=k^2=ijk=x, x^2=1\rangle$ (Note: I am writing $x$ for $-1$, as later I will write $c^x=c$ and if I wasn't to do this I would be writing $c^{-1}=c$ which is silly...). Any non-trivial subgroup of $Q_8$ contains $x$, and so $x$ is contained in the kernel of any map $Q_8\twoheadrightarrow\mathbb{Z}_2$. The image of $Q_8/\langle x\rangle$ is the Klein $4$-group, which maps onto the cyclic group of order two in three ways. Therefore, there are three choices for $\phi$.

In these three automorphisms you always have $c^{x}=c$, and then $c^p=c^{-1}$ for $p\in\{i, j, k\}$ which wasn't killed by our map to the cyclic group of order two, and $c^p=c$ otherwise. The three groups are the following. $$\begin{align*} \langle i,j,k,x, c&; i^2=j^2=k^2=ijk=x, x^2=1, c^2=1, c^x=c, c^i=c^{-1}, c^j=c^{-1}, c^k=c\rangle\\ \langle i,j,k,x, c&; i^2=j^2=k^2=ijk=x, x^2=1, c^2=1, c^x=c, c^i=c^{-1}, c^j=c, c^k=c^{-1}\rangle\\ \langle i,j,k,x, c&; i^2=j^2=k^2=ijk=x, x^2=1, c^2=1, c^x=c, c^i=c, c^j=c^{-1}, c^k=c^{-1}\rangle \end{align*}$$

user1729
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  • $-1$ is contained in any non-trivial subgroup of $Q_8$, so it must be killed if you map to a group of order $2$. – Tobias Kildetoft Sep 09 '13 at 19:46
  • @Tobias I thought that there would be a nice reason (I was too busy trying to prove that the three groups are isomorphic. Which, so far as I can see, they aren't.) I'll edit that tit-bit in, if you don't mind? – user1729 Sep 09 '13 at 19:48
  • @user1729:+1 Plz let me reflect. Thanks – Mikasa Sep 09 '13 at 19:51
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    The first group certainly contains a misprint, I guess you mean $c^{k} = c$. – Andreas Caranti Sep 09 '13 at 21:37
  • @AndreasCaranti Yes, thanks. – user1729 Sep 10 '13 at 09:29
  • @user1729 A very late comment: these 3 groups are isomorphic since subgroups of order 4 in $Q_8$ are all isomorphic to $\mathbb Z_4$, so choose different $\text{ker}(Q_8 \to \mathbb Z_2)$ is equivalent to rename generators of $Q_8$. And $\mathbb Z_3 \rtimes Q_8$ is Dicyclic group – Andrews Nov 08 '19 at 22:50
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It depends on $n$, really. Note that if $n = 2^{m}$, with $m \ge 3$, then $\operatorname{Aut}(G)$ is not cyclic, where $G = \mathbb{Z}_{n}$. Rather, it is the direct product of a cylic group of order $2$, and a cyclic group of order $2^{m-2}$.

So for instance when $n = 8$ you have, besides the groups similar to those mentioned in another answer, a group where $$ c^{i} = c^{-1}, c^{j} = c^{3}, c^{k} = c^{-3}, $$ and variations.

Andreas Caranti
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  • (+1) for your post. Is this a right sample: $$\mathbb Z_3\rtimes _{\phi}Q_8 ;=; \langle a,b,c \mid a^4=1, a^2=b^2, ba=a^3b, c^3=1, ca=ac^2,cb=bc^2\rangle$$ – Mikasa Sep 10 '13 at 08:47
  • Thanks @BabakS., yes, it is one of the groups mentioned in the other answer. – Andreas Caranti Sep 10 '13 at 08:49
  • @BabakS., sorry, I do not understand what you mean, please clarify. – Andreas Caranti Sep 10 '13 at 08:55
  • I am in a doubt about the power $3$ or $-3$ for $c$. Indeed, I thought abot $ca=ac^7$ or $cb=bc^7$. My mind is missing and not to see something clear. !! Sorry. – Mikasa Sep 10 '13 at 08:57
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    @BabakS., my answer was for the case of $\mathbb{Z}_{8}$, whose automorphism group is ${ c \mapsto c, c \mapsto c^{3}, c \mapsto c^{5}, c \mapsto c^{7} }$, where of course you can replace $5$ by $-3$ and $7$ by $-1$, as $c$ has order $8$. – Andreas Caranti Sep 10 '13 at 09:28
  • Thank you for your immense patience. – Mikasa Sep 10 '13 at 09:50