# How many (non-isomorphic) groups of order 315 are there?

Show that all such groups are a direct product of a group of order 5 with the semi direct product of a group of order 7 with a group of order 9. Please help me fix my solution! Help appreciated.

The number $$315=3^2 \cdot 5 \cdot 7$$. We show that the group is a direct product of $$\mathbf{Z}/5\mathbf{Z}$$ with a semi-direct product of a cyclic group of order 7 with a group of order 9.

Using the Sylow theorems, the number of groups of order $$5$$ is either $$21$$ or $$1$$. The number $$n_5= [G:N(P_5)]$$, where $$P_5$$ denotes the Sylow group of order $$5$$. If $$n_p=21$$, then $$|N(P_5)|=3$$, which is not possible since it contains a subgroup $$P_5$$. Therefore, there is a unique group of order 5.

Now we need to determine the number of Sylow-7 subgroups. The number $$n_7 \equiv 1 \mod 7$$ and $$n_7 | 45$$. The only two options are therefore $$15$$ and $$1$$. Let's eliminate 15: we also have by the Sylow theorems that if a Sylow-7 group is called $$P$$, then $$n_p = [G: N_G( P)]$$. If $$n=15$$, then the normalizer has order 23. The group $$P$$ is a subgroup inside of its own normalizer, but $$15$$ does not divide $$23$$.

Now we investigate the groups of order $$9$$. It is easy to show that all groups of order $$p^2$$ are abelian. Therefore, the groups of order $$9$$ are $$\mathbf{Z}/3\mathbf{Z}\times \mathbf{Z}/3\mathbf{Z}$$ and $$\mathbf{Z}/ 9\mathbf{Z}$$. The automorphisms of $$\mathbf{Z}/9\mathbf{Z}$$ are determined by the location of the generator. There are $$\varphi(9) = 6$$ possible locations. There are $$2$$ automorphisms of $$\mathbf{Z}/3\mathbf{Z}$$, plus we can interchange $$(0,1)$$ and $$(1,0)$$ in $$\mathbf{Z}/3\mathbf{Z} \times \mathbf{Z}/3\mathbf{Z}$$, making for a total of $$8$$ possible automorphisms. Subgroups of order 7 have all of their elements order 7, so there cannot possibly be any non-trivial maps from $$\mathbf{Z}/7\mathbf{Z}$$ into these automorphism groups.

Thus, we find that all of the groups of order $$315$$ are abelian and take the form $$\mathbf{Z} /3\mathbf{Z}\times \mathbf{Z}/3\mathbf{Z} \times \mathbf{Z}/7\mathbf{Z}\times \mathbf{Z}/5\mathbf{Z}$$ or $$\mathbf{Z}/9\mathbf{Z}\times \mathbf{Z}/7\mathbf{Z}\times \mathbf{Z}/5\mathbf{Z}$$.

• According to the OEIS there are $4$. Aug 23, 2020 at 19:47
• Your third paragraph is mistaken. If $n_5=21$, then $\vert N(P_5) \vert = 15$. Aug 23, 2020 at 19:52
• Note that there is a nonabelian group of order $21$ (the smallest nonabelian group of odd order) so the abelian groups cannot be the only ones here (since this group can obviously appear as a direct factor) Aug 23, 2020 at 19:55
• Rough outline: the $5$ always lies in the centre, and is a direct factor. The $7$ is always normal. The $9$ is either $3\times 3$ or $9$, and either acts trivially or non-trivially on the $7$. This yields four groups. (This obviously needs proof!) Aug 23, 2020 at 20:14

If $$n_p=21$$, then $$|N(P_5)|=3$$, which is not possible since it contains a subgroup $$P_5$$.

No, if $$n_5 = 21$$, then $$\lvert N(P_5) \rvert = \lvert G \rvert / n_5 = 315/21 = 15$$, which is not obviously impossible.

If $$n=15$$, then the normalizer has order 23.

No, if $$n_7 = 15$$, then the normalizer has order $$315/15 = 21$$.

The group $$P$$ is a subgroup inside of its own normalizer, but $$15$$ does not divide $$23$$.

The order of $$P$$ is $$7$$, not $$15$$. $$7$$ does divide $$21$$.

There are $$2$$ automorphisms of $$\mathbf{Z}/3\mathbf{Z}$$, plus we can interchange $$(0,1)$$ and $$(1,0)$$ in $$\mathbf{Z}/3\mathbf{Z} \times \mathbf{Z}/3\mathbf{Z}$$, making for a total of $$8$$ possible automorphisms.

Actually, there are $$48$$ automorphisms of $$\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$$.

There is a bigger problem here: you seem to be looking at automorphism groups to determine the possibilities for a semidirect product, but you haven't even shown that this group does decompose as a semidirect product! This is easy if all the Sylow subgroups are normal (e.g. by Schur-Zassenhaus), but in general the Sylow subgroups are not all normal (indeed there are non-abelian groups of order 315).