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Let $n$ be a cyclicity-forcing number. Prove that $n=p_1p_2\dots p_r$ is a product of distinct primes and $p_i\nmid p_j-1$ for all $i,j$. [If $n$ is not of this form, construct noncyclic groups of order $n$ using direct products of noncyclic groups of order $p^2$ and $pq$, where $p\mid q-1$.] (Abstract Algebra: Dummit & Foote, Sylow's theorem, Ex. 55)

Note that this is the 'converse' of the classification of cyclicity-forcing number in a sense that there is no other numbers which force the cyclicity of groups of their orders.

Let's say $p_1$ appears twice in the prime factorization of $n$. Write $n=p_1^2p_2\dots p_r$. Then $G\cong\mathbb Z_{p_1}\times\mathbb Z_{p_1}\times\mathbb Z_{p_2}\times\dots\times\mathbb Z_{p_r}$ is a noncyclic group of order $n$. Same if $p_1$ appears more than twice. Just write $k$ $\mathbb Z_{p_1}$'s in the direct product if $p_1$ appears $k$ times in the prime factorization of $n$. The resultant direct product can't be cyclic.

But how can I construct noncyclic groups of order $n$ if $n=p_1p_2\dots p_r$ and $p_i\mid p_j-1$ for some $i,j$? This reminds me of something called a 'semi-direct product'. I peeped through the later texts of the book so I kind of know such thing exists, and I remember that $\operatorname{Dih}(n)$ is a 'semi-direct product' of $\mathbb Z_2$ and $\mathbb Z_n$. But that's it: I know nothing more. Then I found this:

https://en.wikipedia.org/wiki/List_of_small_groups

I noticed that there are only two groups of order $21$ up to isomorphism: $\mathbb Z_7\times\mathbb Z_3$ and $\mathbb Z_7\rtimes\mathbb Z_3$ (Note that $3\mid (7-1)$). How can I 'present' $\mathbb Z_7\rtimes\mathbb Z_3$ and 'construct' noncyclic groups of order divisible by $21$? In general, how can I construct noncyclic groups of order $n$ using direct products of noncyclic groups of $pq$, where $p\mid q-1$, without even introducing the notion of a 'semi-direct product'?

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As you said, by Sylow there are exactly $2$ non-isomorphic groups of order $pq$ with primes $p<q$ and $q\equiv 1 \bmod p$, namely the cyclic one $C_{pq}$, and the non-abelian one, which is the semidirect product $C_q\rtimes C_p$. The good news is, that we need not introduce the semidirect product for this non-abelian group. It is also given as the following matrix group $$ \left\{ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \mid a\in (\mathbb{Z}/(q))^{\times}, b\in \mathbb{Z}/(q), a^p\equiv 1 \bmod q \right\} \cong C_q\rtimes C_p. $$

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  • $\begingroup$ ...but who would have thought of that when attempting this question. I mean, how do you even know that this group is of order $pq$? $\endgroup$ – user441558 Jul 4 '17 at 13:29
  • $\begingroup$ Well, $b$ has $q$ choices, and $a$ has $p(q-1)/(q-1)=p$ choices. And from the example of the affine group one knows that semidirect products and such matrices are very much related. $\endgroup$ – Dietrich Burde Jul 4 '17 at 13:35

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