Let $\bar{m}$ have exponent $k$ in $Z^x_n$, and let $\alpha : Z_k \rightarrow Z^x_n$ be the homomorphism that takes the generator to $\bar{m}$. Then writing $Z_n = \langle b \rangle$, $Z_k = \langle a \rangle$, and identifying $Z^x_n$ with $\operatorname{Aut}(Z_n)$, we obtain

$$Z_n \rtimes_{\alpha} Z_k = \{b^ia^j | 0 \leq i < n, 0 \leq j < k\},$$ where b has order n, a has order k, and the multiplication is given by $$b^ia^jb^{i'}a^{j'} = b^{i+m^ji'}a^{j+j'}.$$

I am a bit confused with the phrase "identifying $Z^x_n$ with $Aut(Z_n)$". What exactly does that mean?

Thanks in advance

  • $\begingroup$ By "Let $\bar{m}$ have exponent k in $Z^x_n$" you meant $\,m\,$ is an element of order $\,k\,$ in the group $\,\Bbb Z_n^*\,$ ? $\endgroup$ – DonAntonio Mar 17 '13 at 12:46
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    $\begingroup$ The group isomorphism $\,\operatorname{Aut}(\Bbb Z_n)\cong \Bbb Z_n^*\,$ is, apparently, what confuses you. $\endgroup$ – DonAntonio Mar 17 '13 at 12:48
  • $\begingroup$ @DonAntonio The textbook defines "exponent" as "Let G be a group and let $n \in Z$. We say that $g \in G$ has exponent n if $g^n=e$". So it's not necessarily the order. $\endgroup$ – user58289 Mar 17 '13 at 12:50
  • $\begingroup$ Please note that $\LaTeX$ editing I did to the text of your question. $\endgroup$ – Andreas Caranti Mar 17 '13 at 12:51
  • $\begingroup$ @Artus, what book is that, please? I think some minimality requirement could probably be considered there... $\endgroup$ – DonAntonio Mar 17 '13 at 12:54

Your $\Bbb{Z}_{n}^{x}$ is really $\Bbb{Z}_{n}^{\star}$, the group (under multiplication) of invertible elements of $\Bbb{Z}_{n}$.

Now each automorphism of the (multiplicatively written) group $\Bbb{Z}_{n} = \langle b \rangle$ is indeed of the form $b \mapsto b^{s}$, for $\bar{s} \in \Bbb{Z}_{n}^{\star}$.

Moreover, the map $$ \Bbb{Z}_{n}^{\star} \to \operatorname{Aut}(\Bbb{Z}_{n}) \qquad \bar{s} \mapsto (b \mapsto b^ {s}) $$ is an isomorphism of groups, that is, the required identification.

  • $\begingroup$ Thanks a lot. But, for example, for $Z_4$, we have $Z_4 = \{0, 1 ,2, 3\}$ and $Z^x_4 = \{1, 3\}$. But then for $b \rightarrow b^s$, s is only allowed to be 1 or 3, right because that's what's in $Z^x_4$? $\endgroup$ – user58289 Mar 17 '13 at 13:08
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    $\begingroup$ @Artus, in your post you were using a multiplicative notation for $\Bbb{Z}_{n}$, so I used it as well. Now you are using additive notation. So you are right that $s = 1, 3$ here, only the automorphism is now $x \mapsto x s$. $\endgroup$ – Andreas Caranti Mar 17 '13 at 13:18
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    $\begingroup$ @Artus, yes, but then you wrote $Z_n = \langle b \rangle$, and wrote powers $b^{j}$. $\endgroup$ – Andreas Caranti Mar 17 '13 at 13:37
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    $\begingroup$ @Artus, you used addition when you wrote, in your first comment, $\Bbb{Z}_{4} = \{0, 1, 2, 3\}$. So here we are talking $\Bbb{Z}_{4}$ as the integers modulo $4$. As an additive group, this is generated wither by $1$ or by $3$. In the question you are (also) using $\Bbb{Z}_{4}$ to denote an abstract cyclic group of order $n$, multiplicatively written, and generated by an element $b$. $\endgroup$ – Andreas Caranti Mar 17 '13 at 13:53
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    $\begingroup$ What we do know is that if $\langle b \rangle \to \Bbb{Z}_{4}$ is an isomorphism, then $b$ is mapped to $1$ or $3$. $\endgroup$ – Andreas Caranti Mar 17 '13 at 14:00

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