Find all non-abelian groups of order 105.

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:

$<x,y |x^{15}=y^7=1, yxy^{-1}=x^\alpha >$

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:

$<x,y |x^{21}=y^5=1, yxy^{-1}=x^\beta >$

But what $\beta$ are possible?

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Hint: if $yxy^{-1} = x^{\alpha}$, then $y^2 x y^{-2} = x^{\alpha^2}$, and... – Qiaochu Yuan May 30 '12 at 21:19
Your first error is that $7 \equiv 1$ mod 3... – Alastair Litterick May 30 '12 at 21:22
According to the small groups library, there are exactly two groups of order 105. As $105=3\cdot5\cdot7$, there is exactly one abelian group of this order, and so there is only one non-abelian group. – jwodder May 30 '12 at 21:28

Let $\,Q_p\,$ be a Sylow p-sbgp. of $\,G\,$ . If either $\,Q_5\,,\,Q_7\,$ is normal then the product $\,P:=Q_5Q_7\,$ is a sbgp. of $\,G\,$ and thus it is normal as well, as its index is the minimal prime dividing $\,|G|\,$. In this case, and since the only existing group of order $35$ is the cyclic one, we'd get that $\,|\operatorname{Aut}(P)|=\phi (35)=24$, and thus we can define a homomorphism $\,Q_3=\langle c\rangle\to\operatorname{Aut}(P=\langle d\rangle)\,$ by $\,d^c:= c^{-1}dc\,$ . As this homom. isn't trivial (of course, we assume at least one of the Sylow sbgps. is non-abelian) we get a (non-abelian, of course) semidirect product $$Q_3 \rtimes P$$ .
If both $\,Q_5,Q_7\,$ are not normal, a simple counting argument tells us there are $\,90\,$ elements of order $7$ and $84$ elements of order 5, which is absurd as we've only $\,105\,$ elements in the group.