# (Dummit and Foote) Group of order 105 with $n_3 = 1$ must be abelian

I was working on this problem: Let $$G$$ be a group of order $$105 = 3\times 5\times 7$$. Assume it has a unique normal Sylow 3-subgroup. Then prove that $$G$$ is abelian.

I worked out the following from Sylow's theorem:

• $$n_5 = 1$$ or $$n_5 = 21$$
• $$n_7 = 1$$ or $$n_7 = 15$$

and by showing that a homomorphism from $$G$$ into $$\operatorname{Aut}(P_q)$$ must be trivial if $$q-1$$ is coprime to $$|G|$$:

• Since $$n_3 = 1$$ the Sylow 3-group lies in the center.
• if $$n_5 = 1$$ the Sylow 5-group lies in the center.

and counting elements of order $$q$$:

• $$n_5 = 21$$ would mean the Sylow 5-subgroups contribute 84 elements of order 5.
• $$n_7 = 15$$ would mean the Sylow 7-subgroups contribute 90 elements of order 7.

This implies we cannot have both, one of them must be a unique normal subgroup.

Is this correct so far? How can I continue from here and finish the proof? Is there a way to avoid splitting into two different cases?

• It's easier. What do you know about $G/P_3$? – Daniel Fischer Aug 7 '20 at 13:38
• In particular, if $x$ centralizes $y$ then $y$ centralizes $x$. – David A. Craven Aug 7 '20 at 13:39
• While it's perfectly allowed to use $G/P_3$, the right strategy concludes the proof in one line, without looking at quotients or knowing anything about the other groups. – David A. Craven Aug 7 '20 at 13:41
• $|G/P_3| = 35$ must be a cyclic group, and is it true that if a quotient of a group by a subgroup of its center is cyclic implies the group is abelian? – river Aug 7 '20 at 13:43
• @JyrkiLahtonen Shall I do a bumper 'how many ways can you prove this?' edit to my solution like here: math.stackexchange.com/questions/3772351/… Then all future 105 questions can be referred to this one. – David A. Craven Aug 7 '20 at 15:08

## 1 Answer

In the hope that this will be the definitive answer to understanding groups of order $$105$$, I will talk about the ways to solve this.

The question assumes that the Sylow $$3$$-subgroup is normal in $$G$$. The condition on the Sylow $$3$$-subgroup here is necessary. There are two groups of order $$105$$, both with a normal Sylow $$5$$- and $$7$$-subgroup, but one is cyclic and the other is $$C_5\times F_{21}$$, where $$F_{21}$$ is a non-abelian group, the normalizer of a Sylow $$7$$-subgroup of $$A_7$$.

The fastest way to proceed is to notice that $$P_3$$, the Sylow $$3$$-subgroup, is not only normal but central. To see this, you can recall that $$N_G(H)/C_G(H)$$ is isomorphic to a subgroup of $$\mathrm{Aut}(H)$$, which has order $$2$$ in this case. The from scratch method is to notice that $$C_3$$ has only two non-identity elements, so for any element $$g\in G$$, $$g^2$$ must act trivially on $$P_3$$. But $$|G|$$ is odd, so every element is a square, and $$P_3$$ is central.

At this point, there are two ways to proceed. The first is to notice that $$G/P_3$$ has order $$35=5\times 7$$, and groups of order $$35$$ are cyclic. If $$G/Z(G)$$ is cyclic then $$G$$ is abelian, and we are done. (Clearly $$G$$ is therefore in fact cyclic.)

The alternative proof is to note that $$P_3\leq C_G(P_5)$$ and $$P_3\leq C_G(P_7)$$. Thus $$|C_G(P_5)|\geq 15$$, and $$|C_G(P_7)|\geq 21$$. (Recall that $$C_G(P_q)\leq N_G(P_q)$$ and $$n_q$$, the number of Sylow $$q$$-subgroups, is equal to $$|G:N_G(P_q)|$$.) From Sylow's theorem ($$n_q\equiv 1\bmod q$$) we see that $$n_5=n_7=1$$, as needed.

If you don't want to do this you can count elements, although it's more subtle than most such arguments. Let's do this without the assumption that $$n_3=1$$, to obtain the full classification.

The number $$n_5$$ of Sylow $$5$$-subgroups is either $$1$$ or $$21=3\times 7$$. We want to prove the former, so assume the latter. Then there are $$21\times 4=82$$ elements of order $$5$$, and since $$C_G(P_5)=P_5$$, there are no elements of order $$5n$$ for any $$n>1$$. This leaves exactly $$105-82=23$$ elements of order not $$5$$, and these must have order $$1$$, $$3$$, $$7$$ or $$21$$. If $$n_7\neq 1$$ then $$n_7=15$$, but this is impossible as there are only $$23$$ elements left. So $$n_7=1$$, removing six elements of order $$7$$. There are seventeen elements left, so $$n_3\leq 8$$ (as each Sylow $$3$$-subgroup requires two elements of order $$3$$). Thus $$n_3=1$$ or $$n_3=7$$. If $$n_3=7$$ then that removes fourteen elements of order $$3$$, and the identity, so there are two elements remaining, which must have order $$21$$. But in any cyclic group of order $$21$$ there are twelve elements of order $$21$$, which is too many.

Thus $$n_3=1$$, and the Sylow $$3$$- and $$7$$-subgroups are both normal. Thus $$P_3P_7$$ is normal in $$G$$, has index $$5$$, and therefore contains every element of order dividing $$21$$. So where are the two remaining elements? This yields a contradiction, so $$n_5=1$$.

If $$n_7\neq 1$$ then $$n_7=15$$, as it must be $$1$$ modulo $$7$$. Again, you can obtain a contradiction as before, because $$C_G(P_5)$$ contains $$P_7$$ but $$C_G(P_7)$$ does not contain $$P_5$$. Let's try to count elements, and see what goes wrong. This yields $$15\times 6=90$$ elements of order $$7$$. There are five elements in $$P_5$$, leaving ten elements. Thus $$n_3\leq 5$$, so $$n_3=1$$. Thus we have a subgroup $$P_3P_5$$ of order $$15$$. This contains ten more elements (as we have already counted $$P_5$$), and so we have exactly the right number of elements, $$105$$.

If $$15$$ were a prime, then this would be fine. Then $$7\mid (15-1)$$ and there would be a map from $$C_7$$ into $$\mathrm{Aut}(C_{15})$$, which would have order $$14$$. But $$15$$ is not a prime, so we can obtain a contradiction using centralizers, as above, but element counting will not work in this case. The group $$P_3P_5$$ has normal subgroups $$P_3$$ and $$P_5$$, on which $$P_7$$ cannot act. Thus $$P_3P_5$$ is actually central, and $$G/(P_3P_5)$$ is cyclic, so $$G$$ is abelian. Alternatively, $$P_3$$ is central, so $$P_3$$ centralizes $$P_7$$. But $$n_7=15$$, so $$P_7$$ does not centralize $$P_3$$. This is a clear contradiction.

Thus $$n_7=1$$ as well. The subgroup $$P_5P_7$$ is a normal, cyclic, subgroup of order $$35$$. Since there is no map from $$P_3$$ to $$\mathrm{Aut}(P_5)$$, this is actually central. The subgroup $$P_7P_3$$, of order $$21$$, complements this, so $$G\cong P_5\times P_7P_3$$. If $$n_3=1$$, equivalently $$P_3$$ centralizes $$P_7$$, then you end up with an abelian (cyclic) group of order $$21$$. If $$n_3=7$$, equivalently $$P_3$$ acts non-trivially on $$P_7$$, then $$P_3P_7$$ is a Frobenius group of order $$21$$. This is the normalizer in $$A_7$$ of a Sylow $$7$$-subgroup.

• Could you tell me why 3 dividing the centralizer implies 3 does not divide n_q? Thanks a lot! – river Aug 7 '20 at 13:53
• Because $n_q=|G:N_G(P_q)|$ and $C_G(P_q)\leq N_G(P_q)$. – David A. Craven Aug 7 '20 at 13:54