TITLE: To construct a non-abelian group of order $55$ and $203$
QUESTION [10 upvotes]: Construct a non-abelian group of order 
i) $55$
ii)$203$
For (i)
I considered $G$, a cyclic group of order $11$ i.e $G$ consists of all $a^i$ where we assume $a^{11}=e$. The mapping $\phi:a^{i} \to a^{4i}$ is an automorphism of $G$ of order $5$ since $\phi^{5}(a^i)=a^{1024i}=a^{1023i}a^{i}=a^{i}.$ Let $x$ be a formal symbol which we subject to the following conditions:
a)$x^{5}=e$
b)$x^{-1}a^{i}x=a^{4i}$
c)$x^{i}a^{j}=x^{k}a^{l}$ if and only if $i \equiv k \pmod 5 $ and $j\equiv l \pmod {11}$
Now we consider all formal symbols $x^{i}a^{j}$ where $i=0,1,2,3,4$ and $j=0,1,...10$. This will be a group, non-abelian of order $55$.
We can folloe similar analysis for (ii).
Is this alright??
Thankyou!!

REPLY [8 votes]: Yes, this is correct.
In general, if $p$ and $q$ are primes with $p$ dividing $q-1$, then
you can find an automorphism of the cyclic group $C_q$ of order $p$,
and then form the corresponding semidirect product of $C_p$ by $C_q$
to obtain a non-abelian group of order $pq$.