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This is a question from an old exam. There should be $4$ groups of order $2014.$ Note $2014 = 2 · 19 · 53$. Admittedly there is an answer Can only find 2 of the 4 groups of order 2014? but I can't make much sense out of the attempt nor the answer.

By Sylow's Theorem, there is a subgroup isomorphic to $\mathbb{Z}/19\mathbb{Z}×\mathbb{Z}/53\mathbb{Z}$. The hint is to observe that conjugation by an element of order two induces an order two automorphism of this subgroup. I am not sure how to proceed. Should the answer depend on the factorization or should a solution be able to work for groups of any order?

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  • $\begingroup$ Hint Well, what are the order $2$ automorphisms of that product? $\endgroup$ – Travis Willse Jun 4 '15 at 10:41
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    $\begingroup$ @Travis that's what I don't understand. I don't recall doing a problem like this before, so I am not sure how to proceed. How would I determine the order 2 automorphisms? $\endgroup$ – cap Jun 5 '15 at 0:04
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Put $n = 2014 = 2.19.53$. By the Sylow theorems, the number $n_p$ of Sylow $p$-subgroups of $G$ has $n_p|(n/p)$ and $n_p \equiv 1\pmod{p}$ for $p = 2, 19, 53$. It follows that $n_p = 1$ for $p = 19, 53$; let $C_p \simeq \mathbb{Z}/p\mathbb{Z}$ denote the Sylow $p$-subgroup of $G$. Clearly $C_{19}\cap C_{51} = 1$, so the group $K = C_{19} C_{51} \subset G$ is isomorphic to $\mathbb{Z}_{19} \times \mathbb{Z}_{51}$. Furthermore, the groups $C_{19}, C_{51}$ are normal in $G$ by the Sylow theorems, so $K$ is as well. For any $g\in G$ of order $2$ (which must exist, by the Sylow theorems), we have $\langle{g, K\rangle} = G$ with $g$ acting on $K$ by conjugation. Thus $G$ is a semidirect product $\langle{g\rangle}\ltimes K = \mathbb{Z}_2 \ltimes (\mathbb{Z}_{19}\times \mathbb{Z}_{51})$ with respect to some action $\mathbb{Z}_2 \to \operatorname{Aut}(\mathbb{Z}_{19}\times \mathbb{Z}_{51})$. (If you're not familiar with semidirect products, I'd recommend you look up the construction; they commonly pop up in problems like this, and it's easier to work from the definition rather than trying to describe it ad hoc.)

So, now we need to compute $\operatorname{Aut}(\mathbb{Z}_{19}\times \mathbb{Z}_{51}) = \operatorname{Aut}(\mathbb{Z}_{n/2})$. The automorphism group of a cyclic group $\mathbb{Z}_m$ is clearly simply $\mathbb{Z}_m^{\times}$, comprising the maps $g \to g^k$ with $(k, m) = 1$ for some fixed generator $g\in \mathbb{Z}_m$. The group of maps $\mathbb{Z}_2 \to \operatorname{Aut}(\mathbb{Z}_{19} \times \mathbb{Z}_{51})$ thus comprises the four maps sending a generator to $(g, h) \to (g^{\alpha}, h^{\beta})$, where $\alpha^2 = 1$ in $\mathbb{Z}_{19}$ and $\beta^2 = 1$ in $\mathbb{Z}_{51}$. This construction gives the required four groups; to show that they're distinct, consider the centralizer of an element of order $2$.

The same approach works in general, though it depends on having $n$ be the product of convenient primes. The technique is honestly not that powerful, but this kind of computation appears on a test in every single class I've seen covering the Sylow theorems, and even some candidacy exams. In this particular case, you're lucky that the divisors of $n$ are reasonably spaced primes, so that the combinatorial constraints imposed by the Sylow theorems are so limiting; this sort of problem often involves some extra complications with solvability, prime powers in the order of $G$, etc.

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  • $\begingroup$ What does $<g,K>$ stand for? What is $n$ in $\mathbb{Z}_{n/2}$? $\endgroup$ – cap Jun 7 '15 at 5:59
  • $\begingroup$ The former is the group generated by $g$ and $K$. The latter is $n = 2014$. $\endgroup$ – anomaly Jun 7 '15 at 6:27
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For cyclic group $C_p$ with $p$ being a prime number, the endomorphism ring $\text{End}(C_p)$ is the field $\mathbb Z_p$, so the automorphism group $\text{Aut}(C_p)$ is the unit group $(\mathbb Z_p)^\times$ of $\mathbb Z_p$, which is isomorphic to $C_{p-1}$. Thus we get

$\text{Aut}(C_{19}\times C_{53})\cong\text{Aut}(C_{19})\times\text{Aut}(C_{53})\cong(\mathbb Z_{19})^\times\times(\mathbb Z_{53})^\times\cong C_{18}\times C_{52}\cong C_{2}\times C_{9}\times C_{4}\times C_{13}$,

we see clearly there are $4$ elements of order at most $2$.

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  • $\begingroup$ Why are you factoring some of the integers but not the others? For example, you write $18$ as $2\times 9$, but then don't factor $9$. What are the 4 groups. Where is the "order 2" coming from? $\endgroup$ – cap Jun 7 '15 at 5:53
  • $\begingroup$ @cap Because given a cyclic group $C_{mn}$, you can only factor it as $C_m\times C_n$ when $m,n$ are coprime, otherwise $C_{mn}$ and $C_m\times C_n$ are not isomorphic. $\endgroup$ – Censi LI Jun 7 '15 at 6:17
  • $\begingroup$ @cap "order 2" comes form the subgroup $C_2\times C_4$ $\endgroup$ – Censi LI Jun 7 '15 at 6:19
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Note that $\mathbb Z/19\mathbb Z\times \mathbb Z/53\mathbb Z\cong \mathbb Z/1007\mathbb Z$. An automorphism of $\mathbb Z/n\mathbb Z$ is of the form $x\mapsto ax$ where $a$ is relatively prime to $n$. For it to be order $2$, then, you need $a^2=1$. So you need to find the square roots of $1$ in the ring $\mathbb Z/1007\mathbb Z$. Chinese remainder theorem means that $a\equiv \pm 1\pmod{19}$ and $a\equiv \pm 1\pmod{53}$.

I get $a\equiv \pm 1, \pm 476\pmod{1007}$.

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  • $\begingroup$ how does one find the group corresponding to one of your $a$ values? $\endgroup$ – cap Jun 8 '15 at 5:59
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Thomas Andrews' solution is complete, but to solve the problem using a little more group theory: a finite group $G$ with a Sylow $2$-subgroup of order $2$ always has a normal subgroup of index $2$ ( consisting of those elements acting as even permutations in the Cayley representation of $G$ acting as a permutation group on itself by right translation). Any group $Y$ of order $1007 =19 \times 53$ is cyclic (say by Sylow's Theorems) and has a characteristic subgroup $A$ of order $19$ and a characteristic subgroup $B$ of order $53$. If $X$ is our group of order $2014$, then $X = \langle t \rangle Y,$ where $Y = A \times B$ is cyclic of order $1007$ and $A = \langle a \rangle $ and $B = \langle b \rangle $ are as above. There are $4$ (and only $4$) ways that $t$ can acts on $Y$:

  1. $t^{-1}at = a, t^{-1}bt = b$.

  2. $t^{-1} a t = a^{-1}, t^{-1}bt = b$.

  3. $t^{-1} at = a,t^{-1}bt = b^{-1}.$

  4. $t^{-1}at = a^{-1},t^{-1}bt = b^{-1}.$

Each of these different actions gives $X = \langle t \rangle Y$ the structure of a semidirect product ( the first being direct), and these $4$ actions clearly lead to $4$ non-isomorphic groups.

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