# Creating groups with $n$ elements for a given $n$

I would like to create groups with $n$ elements for a given $n$.

I know that I can take some product of $\mathbb Z / p_i \mathbb Z$ for some primes $p_i$. But I want to find less obvious (=more interesting) ways to construct groups.

My ideas so far:

For example, let $n=105$ or $n=44$.

My first idea was to use groups of units, $U(n)$. Then I found out that this cannot work for odd $n$ because the $U(n)$ always have even number of elements. So this already fails for $n=105$.

My second idea was that one could consider matrix groups, like the special linear group. But the number of elements, say over $\mathbb K = \mathbb Z / p \mathbb Z$, is a square number. Of course, neither $105$ or $44$ are square numbers. So this method is no good either.

My third idea was to use some cyclic subgroup of some group, possible the group of units, but this idea is a bit fuzzy as it's not clear to me to construct the right number of elements in the subgroup. Is there a method to define a subgroup that has $n$ elements?

More generally:

What methods are there to construct groups of a given number $n$ of elements?

• Keep in mind that for some $n$'s, the only possibilities are products of $\Bbb Z/ p\Bbb Z$. – Daniel Oct 28 '15 at 5:54
• @SolidSnake I did not know this. How do I know which $n$s? – a student Oct 28 '15 at 5:55
• For example, if $n$ is prime, then the only group of order $n$ is $\Bbb Z/ n\Bbb Z$ (this comes from Lagrange's theorem). – Daniel Oct 28 '15 at 5:57
• @astudent These are called cyclic numbers, they are characterized by $n$ and $\phi(n)$ being relatively prime. In concrete terms, $n$ is squarefree and no two primes $p,q$ dividing $n$ satisfy $p \mid q-1$. – Erick Wong Oct 28 '15 at 6:08
• You can take semidirect products. – Qiaochu Yuan Oct 28 '15 at 6:23

There is a nonabelian group of order $21$ given by the group of affine linear transformations $z \mapsto az + b$ over $\mathbb{F}_7$ where $a$ is a nonzero square in $\mathbb{F}_7^{\times}$. Taking the product of this group with $\mathbb{Z}_5$ produces a nonabelian group of order $105$.
Similarly, there is a nonabelian group of order $22$ given by the group of affine linear transformations $z \mapsto az + b$ over $\mathbb{F}_{11}$ where $a = \pm 1$. Taking the product of this group with $\mathbb{Z}_2$ produces a nonabelian group of order $44$ (I think there are exactly two such groups).