2

I’m trying to prove that if $G$ is a group of order $90$, then it contains a subgroup of order $10$. This is what I have so far:

“The prime factorization of $90$ is $3^2 \cdot 2 \cdot 5$. By the First Sylow Theorem, Sylow $p$-subgroups exist, meaning that there is at least one Sylow $2$-subgroup, say $P$ and there is at least one Sylow $5$-subgroup, say $Q$. Additionally, Sylow $p$-subgroups intersect trivially, meaning that $|P \cap Q|=1$.
Therefore, we have $|PQ|=\frac{|P||Q|}{|P \cap Q|}=\frac{2 \cdot 5}{1}=10$.
And so, $G$ contains a subgroup $PQ$ of order $10$, as was to be shown.”

Is this correct?

Shaun
  • 47,747
MATH-LORD
  • 400
  • 1
    $PQ$ need not be a subgroup unless one of $P$ or $Q$ is normal. – Clement Yung Aug 16 '20 at 17:12
  • 1
    You didn't prove that $PQ$ is indeed a subgroup. It is not always the case. However, it is true if, for example, one of the groups $P$ and $Q$ is normal in $G$. I'll give a hint: prove that $Q$ is the only $5$-Sylow subgroup, hence it must be normal. – Mark Aug 16 '20 at 17:12
  • 1
    By the way, you wrote that "Sylow $p$-subgroups intersect trivially". In your case you intersect a $p$-Sylow subgroup with a $q$-Sylow subgroup where $p\ne q$, in this case the intersection indeed must be trivial. (easily follows from Lagrange's theorem). But I hope you know that if you take two $p$-Sylow subgroups of the same prime $p$ then their intersection might not be trivial. So your explanation to why the intersection is trivial is not good enough. You should add that you are intersecting Sylow subgroups related to different prime divisors of $|G|$. – Mark Aug 16 '20 at 17:16
  • $PQ$ isn’t a subgroup even if $P$ and $Q$ have relatively prime order? – MATH-LORD Aug 16 '20 at 17:25
  • 1
    It isn't enough. If we take $G=S_4, H=\langle (13)\rangle$ and $K=\langle (124)\rangle$ then unless I did any mistakes in calculations $HK$ is not a subgroup. In general, $PQ$ is a subgroup if and only if $PQ=QP$. A sufficient (though not necessary) condition which is usually easier to check is that at least one of the subgroups is normal. – Mark Aug 16 '20 at 17:35
  • 1
    Just try to show this set has a unique $5$-Sylow subgroup. Do you know how to find the number of $p$-Sylow subgroups? – Umesh Shankar Aug 16 '20 at 18:29
  • 2
    @UmeshShankar Easier said than done, as eliminating the $n_5=6$ case looks to be notably non-trivial. Even proving the group can't be simple is kind of non-trivial. – zibadawa timmy Aug 17 '20 at 06:07

3 Answers3

4

While I admire the other poster's persistence, here is a faster way.

Suppose that $n_5=6$. Then $G$ acts on the six Sylow $5$-subgroups. As $|N_G(P)|=15$, and this is the important bit, no element of order $2$ can normalize any Sylow $5$-subgroup. Thus the permutation action of an element of order $2$ must be (up to labelling) $(1,2)(3,4)(5,6)$, an odd permutation. Let $H$ be the set of elements that induce an even permutation on the Sylow $5$-subgroups. Then $|G:H|=2$, and thus all elements of order $5$ lie in $H$. By Sylow's theorem, $n_5=1$ for $H$ (no other option), and therefore $n_5=1$ for $G$, a contradiction.

David A. Craven
  • 11,561
  • 1
    This still relies on the observation crucial to the more general Cayley's application. That $H$ has index 2 turns on the same observation that the more general Cayley argument does: right multiplication by the given odd permutation yields a bijection between even permutations in the image and odd permutations in the image. Still, one way or another, terribly clever. – zibadawa timmy Aug 17 '20 at 10:13
  • 1
    I feel that since that observation was made in 1878 it should be considered part of the standard toolkit! You are right that it's about odd permutations in actions, but this is by conjugation rather than right multiplication, so it's slightly different. This method does not generalize, for example. – David A. Craven Aug 17 '20 at 10:45
2

As noted in the comments, your proof depends on $PQ$ being a subgroup, which need not be true in general. If one normalizes the other ($P\subseteq N_G(Q)$ or $Q\subseteq N_G(P)$) then it works and $PQ$ is a group. This is certainly the case if one of them is in fact normal in $G$. But you haven't shown that $PQ$ is a subgroup, so your proof is incomplete at best. The hope would be to prove that $Q$ is necessarily normal. This turns out to be true, though proving it abstractly, rather than by computer, is rather obscene. The proof is ultimately one of counting elements first to constrain the problem, and then a bunch of subcases to consider with various methods.

Ultimately, I will prove slightly less than that the Sylow 5-subgroup is normal, stopping once we have either a normal Sylow 2-subgroup, a normal Sylow 5-subgroup, or have constructed a subgroup of order 10, or proven the arrangement in question is impossible so can be excluded.

Consider the number of Sylow $5$-subgroups, denoted $n_5$. By the Sylow theorems, $n_5\equiv 1\bmod 5$ and $n_5$ divides $2\cdot 3^2=18$. Thus either $n_5=1$ or $n_5=6$. If $n_5=1$ then your $Q$ is in fact normal and so your $PQ$ is a subgroup of order 10, and we are done. But what to do in the case $n_5=6$? Ultimately this is impossible, but we will show how we can always exhibit a subgroup of order 10 in any case we can't derive a contradiction from.

So suppose $n_5=6$. The conjugation action transitively permutes the 6 Sylow 5-subgroups. By applying the orbit-stabilizer theorem, or one of the Sylow theorems that's just a special case thereof, we have that $[G:N_G(Q)]=n_5=6$, so that $N_G(Q)$ is a subgroup of order 15 and index 6 in $G$. Note that every group of order 15 is cyclic.

Now distinct conjugates of $Q$ have distinct normalizers since $N_G(tQt^{-1})=t N_G(Q) t^{-1}$ for all $t\in G$ (they can intersect non-trivially, but can't be equal), so no two such normalizers share an element of order 15. As a cyclic group of order 15 contains exactly 8 elements of order 15, the normalizers of the conjugates of $Q$ yield a total of 48 elements of order 15 in $G$.

Since the Sylow 5-subgroups are cyclic of order 5, and contain exactly 4 elements of order 5, we have a total of 24 elements of order 5 in $G$.

Combined, we've accounted for 72 non-identity elements in $G$, none of which has order (divisible by) 2.

We now consider the number of elements of order 2. Since the Sylow 2-subgroup is cyclic of order 2, this is precisely the number of Sylow 2-subgroups of $G$, denoted $n_2$. By the Sylow theorems, we have that $n_2\in\{1,3,5,9,15,45\}$, the odd divisors of 90. Our aim is to show that in each case we either have a contradiction, or can exhibit a group of order 10. This then establishes your desired result.

Fix $P\in\operatorname{Syl}_2(G)$.

We first claim that $n_2=[G:N_G(P)]$ is divisible by 5, so that $n_2\in\{5,15,45\}$; it's possible to ignore this and still handle the other values of $n_2$ with relative ease, but it's a convenient reduction. To see this, note that since $P=\langle\,y\,\rangle$ is cyclic of order 2, then in fact $N_G(P)=C_G(y)$. If $N_G(P)$ had order divisible by 5, it would have an element of order 5, which implies that $y$ centralizes an element of order 5. Thus $y$ normalizes some Sylow 5-subgroup, but our assumption that $n_5=6$ implies that the normalizer of a Sylow 5-subgroup has odd order. Thus $n_2$ is divisible by 5, as desired.

If $n_2=45$ then we have accounted for $72+45>90$ non-identity elements in $G$, which is impossible. So $n_2\neq 45$.

If $n_2=15$, then we've accounted for $72+15=88$ non-identity elements, leaving at most 3 non-identity elements for a Sylow 3-subgroup. But since such a subgroup has exactly 9 elements, this is also impossible. So $n_2\neq 15$.

Lastly, we must consider the case $n_2=5$. This implies that $|N_G(P)|=18$. The conjugation action on the Sylow 2-subgroups gives us a group homomorphism $\phi\colon G\to S_5$. This homomorphism would be trivial only if $P$ was normal, which we have assumed is not the case. It also cannot be injective since 90 does not divide $|S_5|=120$. Indeed $9$ does not divide $120$, so the kernel must have order divisible by 3. Since we also must have $\ker(\phi)\subseteq N_G(P)$, we conclude that $\ker(\phi)$ has order in $\{3,6,9\}$ and that $\phi(G)$ contains a cyclic subgroup $C$ of order 5.

We claim that if we know that a group of order 30 has a subgroup of order 10, then we are left with the case $|\ker(\phi)|=9$.

So let us see where the group of order 30 business comes in. If $\ker(\phi)$ has order 6 then $\phi^{-1}(C)$ has order 30. If $\ker(\phi)$ has order 3, then $\phi(G)$ has order 30. If $H$ is then a subgroup of order 10 in $\phi(G)$, then $\phi^{-1}(H)$ is a subgroup of order 30. This establishes the claim.

That groups of order 30 admit a subgroup of order 10 is left as an exercise. You can try much the same thing as before: if the Sylow 5-subgroup isn't normal, then there's 6 of them, 24 elements of order 5, etc. (Or see the much easier proof of this whole problem in my other answer that David A. Craven pointed out, and apply the same argument)

This leaves the case $|\ker(\phi)|=9$ (and $n_2=5$) as the only one left to consider. Then $\phi(G)$ has order 10, but this is a quotient group of $G$, and does not yield a subgroup of $G$ as before. But $\phi(G)$ is necessarily cyclic, so $C$ is normal in $\phi(G)$, and so $\phi^{-1}(C)=L$ is a normal subgroup of order 45 in $G$.

By order considerations, a Sylow 5-subgroup of $L$ is a Sylow 5-subgroup of $G$. Indeed, since $L$ is normal and the conjugation action is transitive on the Sylow 5-subgroups, $\operatorname{Syl}_5(G)=\operatorname{Syl}_5(L)$. We claim that in fact a group of order 45 always has a normal Sylow 5-subgroup, which then gives us a contradiction, and at long last completes all cases and proves that $G$ has a subgroup of order 10.

So, how to prove a group of order 45 has a normal Sylow 5-subgroup? Well, that is, at long last, easy! By the Sylow theorems, the number of Sylow 5-subgroups of such a group is coprime to 5 and divides 9. The only possibility is therefore 1.

QED.

zibadawa timmy
  • 4,533
  • 3
    It's quicker to prove Cayley's theorem that $G$ has a subgroup of index $2$ if $|G|=2n$, $n$ odd, and use that. This proof isway too long, and cannot be what is needed. – David A. Craven Aug 17 '20 at 09:38
  • 1
    @DavidA.Craven You may be right, as the index 2 subgroup does make a helpfully easy appearance at the end, but my mind had already gone deep down the rabbit hole and put on its blinders before it really showed up. – zibadawa timmy Aug 17 '20 at 09:40
  • 1
    If $|G:N_G(P)|=5$ then this yields a homomorphism into $S_5$. Since $9\nmid |S_5|$, the kernel must have order a multiple of $3$. This gives us a normal subgroup of $5'$-order. Quotient out by that, use induction, and see that $n_5=1$. (Of course, $n_2$ is a multiple of $5$ because $n_5=6$, so no element of order $5$ centralizes an element of order $2$.) – David A. Craven Aug 17 '20 at 09:42
  • 1
    @DavidA.Craven Oops, had scribbled down the multiple 3 part on paper then casually ignored it when typing it up. Probably because before I checked the paper I realized the order 2 case was also easy to get the desired subgroup from, which was the objective in mind. But it'd be better to note we can exclude it entirely... – zibadawa timmy Aug 17 '20 at 09:57
2

Following David A. Craven's remark on my more arduous proof:

Here you can find a proof that a group of order 4n+2 has a (normal) subgroup of index 2, using little more than Cayley's theorem to imbed into a symmetric group of degree 4n+2. Now $90=4\cdot 22+2$ is of this form, so a group $G$ of order 90 always has a subgroup of index 2. So we have a normal subgroup $H$ of order 45, and every Sylow 5-subgroup of $G$ is a Sylow 5-subgroup of $H$ and vice versa. But an easy application of the Sylow Theorems shows that a group of order 45 always has a normal Sylow 5-subgroup, so $G$ must have a unique and therefore normal Sylow 5-subgroup.

Your argument then carries through, as your product does in fact define a subgroup.

zibadawa timmy
  • 4,533