Find all non-abelian groups of order 105.
My attempt: $105=3.5.7$.
Consider the $3$-factorization, as $5, 7, 35 \not\equiv 1\mod 3$ we have that the sylow $3$-subgroup is normal.
Consider now the $7$-factorization. By this we get that there is either $15$ sylow $7$-subgroups or one normal sylow $7$-subgroup.
If there are $15$ sylow-$7$-subgroups then $[N_G(P):P]=1$ and then since $P$ is abelian we have that $N_G(P)=C_G(P)$. But this means that we have that P has a normal $p$-complement and so we have that any groups of this form can be represented as:
$
But what $\alpha$ are possible?
If there is one normal sylow $7$-subgroup then we have a normal subgroup of order $21$ which is abelian. Hence we have groups of this type represented by:
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But what $\beta$ are possible?