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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?

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    $\begingroup$ It's easier. What do you know about $G/P_3$? $\endgroup$ Aug 7, 2020 at 13:38
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    $\begingroup$ In particular, if $x$ centralizes $y$ then $y$ centralizes $x$. $\endgroup$ Aug 7, 2020 at 13:39
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    $\begingroup$ 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. $\endgroup$ Aug 7, 2020 at 13:41
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    $\begingroup$ $|G/P_3| = 35$ 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? $\endgroup$
    – user581023
    Aug 7, 2020 at 13:43
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    $\begingroup$ @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. $\endgroup$ Aug 7, 2020 at 15:08

1 Answer 1

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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.

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  • $\begingroup$ Could you tell me why 3 dividing the centralizer implies 3 does not divide n_q? Thanks a lot! $\endgroup$
    – user581023
    Aug 7, 2020 at 13:53
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    $\begingroup$ Because $n_q=|G:N_G(P_q)|$ and $C_G(P_q)\leq N_G(P_q)$. $\endgroup$ Aug 7, 2020 at 13:54

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