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Compute $|\operatorname{Aut} (\mathbb{(Z}/{1155\mathbb Z)}^\times)|$.

I feel puzzled about this problem. I found that $$\mathbb{(Z}/{1155\mathbb Z})^\times \cong \mathbb{(Z}/3\mathbb Z)^\times \times \mathbb{(Z}/5\mathbb Z)^\times \times \mathbb{(Z}/7\mathbb Z)^\times \times \mathbb{(Z}/{11\mathbb Z)}^\times. $$ I get the order of itself $\left|\mathbb{(Z}/{1155\mathbb Z)}^\times\right|$, but I don't know how to find its automorphism.

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    Please format your question properly. – 2017-01-22
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    Sorry，I will learn the latex and pay more attention next time. – 2017-01-22
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    The next thing you should do is probably to write that group in a form where the invariant factors show. Like $$\Bbb{Z}_2\times\Bbb{Z}_4\times\Bbb{Z}_6\times\Bbb{Z}_{10}\simeq\Bbb{Z}_{60}\times\Bbb{Z}_2^3.$$ Do you see how to get there? – 2017-01-22
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    Does $\Bbb{Z}^\times_{1155}$ mean the collection of non-zero elements? – 2017-01-22
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    “https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n”is this mean .but I don not how to type the form correctly on this site via using the code.@Frank Lu,@Jyrki Lahtonen I don‘t know your meaning would you like to teach me more detail about it? – 2017-01-22
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    Jackunir: I wrote the group $\Bbb{Z}_{1155}^*$ as a product of cyclic groups in such a way that the order of a factor always divides those of the preceding ones, or in terms of [invariant factors](https://en.wikipedia.org/wiki/Invariant_factor). I think that it is probably easier to figure out what kind of automorphisms there are, when you write the group in this somewhat canonical form. – 2017-01-22
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    thank you ,(I have learned if the whole group is cyclic,we can use Euler function φ to compute its automorphism group order in my algebra class ,but I don't know how to compute the product groups' order after decomposing the 1155 into the cyclic groups product. Are there exist lemmas to compute product group's automorphism which I have little knowledge about it?)@Jyrki Lahtonen♦ – 2017-01-22
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    Jackunir, probably there are formulas. I just don't remember them, so my method was a bit ad hoc. IIRC Jacobson's *Basic Algebra I, chapter 3* has a formula, but I'm not sure whether it was for automorphisms or endomorphisms. Anyway, the idea to use invariant factors is described there. Probably in other texts on abstract algebra as well. – 2017-01-22
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    thank you，the book what i learnt conclude little about automorphisms... – 2017-01-22

2 Answers 2

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Let us first rewrite the abelian group $G=\Bbb{Z}_{1155}^*$ in terms of its invariant factors. You already deduced that $$ \Bbb{Z}_{1155}^*\simeq \Bbb{Z}_3^*\times\Bbb{Z}_5^*\times\Bbb{Z}_7^*\times\Bbb{Z}_{11}^*. $$ We know that for a prime $p$, $\Bbb{Z}_p^*$ is cyclic of order $p-1$. Therefore $$ G\simeq C_2\times C_4\times C_6\times C_{10}. $$ Recall the rule that if $\gcd(m,n)=1$, then $C_m\times C_n\simeq C_{mn}$. Here $C_6\simeq C_2\times C_3$ and $C_{10}\simeq C_2\times C_5$. Regrouping gives thus $$ \begin{aligned} G&\simeq C_2\times C_4\times(C_2\times C_3)\times(C_2\times C_5)\\ &\simeq (C_4\times C_3\times C_5)\times C_2\times C_2\times C_2\\ &\simeq C_{60}\times C_2\times C_2\times C_2. \end{aligned} $$ IMHO it is easier to count the number of automorphisms of $G$ from this. Let $c_1,c_2,c_3,c_4$ be generators for the four factors, so $c_1$ is of order $60$ and $c_2,c_3,c_4$ are all of order two. We get all the elements of $G$ uniquely in the form $\prod_{i=1}^4 c_i^{a_i}$ with $0\le a_1<60$, $a_2,a_3,a_4\in\{0,1\}$.

Let $\sigma$ be an automorphism of $G$. Then obviously $\sigma(c_1)$ is of order $60$. Therefore we must have $\sigma(c_1)=c_1^{a_1}c_2^{a_2}c_3^{a_3}c_4^{a_4}$ with $\gcd(a_1,60)=1$ but no constraints on the other exponents. There are $\phi(60)=2\cdot(3-1)\cdot(5-1)=16$ choices for $a_1$, and two choices for the other exponents - a total of $16\cdot2^3=128$ choices for $\sigma(c_1)$ (we shall shortly see that all these choices lead to automorphisms of $\sigma$). Let us record the fact that because $a_1$ is necessarily odd we have $\phi(c_1^{30})=c_1^{30}$. Furthermore, this is the only element of order two in $\langle \sigma(c_1)\rangle$ (a cyclic group of an even order has a unique element of order two).

The elements of order two in $G$ are exactly the fifteen non-trivial elements of the elementary 2-abelian subgroup $H$ generated by $\{c_1^{30},c_2,c_3,c_4\}$. Of these $c_1^{30}=\sigma(c_1^{30})$ has already been used. This means that fourteen possible choices remain for $\sigma(c_2)$.

From this point the calculation proceeds very similarly to the calculation of the number of invertible matrices modulo two. Namely, $\sigma(c_3)$ can be any element of $G$ that is of order two, but is not in the subgroup generated by $\sigma(c_1^{30})$ and $\sigma(c_2)$ (otherwise $\sigma$ won't be injective). The latter condition rules out $4$ of the $16$ elements of $H$, and, given $\sigma(c_1)$ and $\sigma(c_2)$, we have $12$ choices for $\sigma(c_3)$.

Repeating the above calculation/argument we see that we have $8$ choices for $\sigma(c_4)$, namely the elements of $H\setminus\langle \sigma(c_1^{30})=c_1^{30},\sigma(c_2),\sigma(c_3)\rangle$. We easily see that all these combinations of choices lead to an injective homomorphism from $G$ to itself, hence to an automorphism. So, finally:

The number of automorphisms of the group $\Bbb{Z}_{1155}^*$ is $$|Aut(G)|=128\cdot14\cdot12\cdot8=2^{13}\cdot3\cdot7.$$

I won't give you the structure of $Aut(G)$ (I don't know how to do that concisely). I do want to point out that $Aut(G)$ obviously has a subgroup isomorphic to the simple group $GL_3(\Bbb{F}_2)$ of order $168$. In particular, $Aut(G)$ is not a solvable group.

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    thank you very much, I'm learning too little about how to solve these problem.would you like to teach me the idea about decomposing it and then recombine asC60×C2×C2×C2 .the aim to recombine C60 is it easy to compute?is there exist some common number that can be our first thinking ？@Jyrki Lahtonen♦(I am a freshman and learning undergraduate course just before this vocation ) – 2017-01-22
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    Very nice (+1). – 2017-01-22
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Hints:

  1. We have $(\mathbb{Z}/p\mathbb{Z})^*\cong \mathbb{Z}/(p-1)\mathbb{Z}$, see here.

  2. We know Aut $\mathbb Z/n\mathbb{Z}$, see here, and this duplicate.

  3. You have obtained already, that $p=3,5,7,11$.

  4. Jyrki's objection for the invariant factors, i.e., to write the direct product of these four cyclic groups in a canonical form.