4
$\begingroup$

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?

$\endgroup$
9
  • 4
    $\begingroup$ It's easier. What do you know about $G/P_3$? $\endgroup$ – Daniel Fischer Aug 7 '20 at 13:38
  • 2
    $\begingroup$ In particular, if $x$ centralizes $y$ then $y$ centralizes $x$. $\endgroup$ – David A. Craven Aug 7 '20 at 13:39
  • 2
    $\begingroup$ 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. $\endgroup$ – David A. Craven Aug 7 '20 at 13:41
  • 1
    $\begingroup$ $|G/P_3| = 35$ 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? $\endgroup$ – river Aug 7 '20 at 13:43
  • 3
    $\begingroup$ @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. $\endgroup$ – David A. Craven Aug 7 '20 at 15:08
5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Could you tell me why 3 dividing the centralizer implies 3 does not divide n_q? Thanks a lot! $\endgroup$ – river Aug 7 '20 at 13:53
  • 1
    $\begingroup$ Because $n_q=|G:N_G(P_q)|$ and $C_G(P_q)\leq N_G(P_q)$. $\endgroup$ – David A. Craven Aug 7 '20 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.