The other day I was reading Topics in Algebra by I.N.Herstein and found a problem which can be summarised as -
Show that any two non-abelian groups of order $pq$ where $p$ and $q$ are distinct primes with $p>q$, are isomorphic.
My strategy
there exists a subgroup of order $p$ and $q$ out of which the $p$-group is a normal group by using the fact that
there exists a homomorphism $\alpha:G\to S_q$ such that the kernel is contained in the $p$-group.
As $p$ and $q$ are cyclic let their generator be $a$ and $b$ respectively. As $\langle a\rangle \cap\langle b\rangle=\phi$ then $\forall g\in G$ $g=a^ib^j$. As $\langle a\rangle $ is normal $bab^{-1}=a^k \implies b^iab^{-i}=a^{k^i} \implies k^q \equiv 1\pmod{p})$ , $k\neq1 $ as it is non-abelian group. Let the two groups be $G_1$ and $G_2$. Let $a_1$ and $b_1$ are the generators of $p$-group and $q$-group respectively and similarly $a_2$ and $b_2$ for $G_2$. Let $$b_1a_1b_1^{-1}=a_1^{k_1}$$ $$b_2a_2b_2^{-1}=a_2^{k_2}$$ If I use the fact that $\Bbb Z_p$ multiplicative group is cyclic, then there exists $n\in N$ $,n\le q-1$ such that $[k_2]^n=[k_1] $ and $[k_1],[k_2] \in \Bbb Z_p$. hence $\beta :G_1 \to G_2$ , $\beta(a_1)=a_2$ and $\beta(b_1)=b_2^n$ is an isomorphism. But the fact that $\Bbb Z_p$ multiplicative group is cyclic is not done in the previous parts. So, I am asking for a proof without using the fact that $\Bbb Z_p$ multiplicative group is cyclic.
