Sylow 3-subgroup and a Sylow 5-subgroup. $|H|=3$ and $|K|=5$. If 3 divides $|N(K)|$, show that 5 divides $|N(H)|$.

Question

The question is:

Let $H$ and $K$ denote a Sylow 3-subgroup and a Sylow 5-subgroup of a group, respectively. Suppose that $|H|= 3$ and $|K|= 5$. If 3 divides $|N(K)|$, show that 5 divides $|N(H)|$.

I have found various hints and solutions ,such as the relation $gN(H)g^{-1}=N(gHg^{-1})$, or hints in here: Sylow $p$-subgroups and their normalizers.

But I do not actually lealize how to complete it. On the other hand,I try the seemly simpler method:

Since 3 divide $|N(K)|$, it has a cyclic subgroup $S$ of order 3.and since $H$ is of order 5 I can write $SH$ of order 15.it is normal in the original group $G$ because it is cyclic, then $N(SH)$ is of order 15t.and I can take $S=\text{identity}$,so $N(H)$ is divided by 15.

I think there is something wrong...,but I cannot find. Any adjustion or advices are appreciate please!

Answer 1

I do not think the line "I can write $SH$ of order 15" is necessarily true. In fact, it is not even clear that $SH$ is a subgroup. Recall that a $HK$ is a subgroup iff $HK = KH$. A corollary to this result is that if $H$ is contained in the normalizer of $K$ (that is $H \leq N(K)$), then $HK$ is a subgroup. See Dummit and Foote section 3.1 proposition 14 and corollary 15 for these results.

However, I think you are on the right track. Try using the relation $gN(H)g^{-1} = N(gHg^{-1})$. Here's one way:

If $3 \mid |N(K)|$, then $N(K)$ contains a sylow 3-subgroup, say $H'$ (note that we don't know if $H$ is a subgroup of $K$). Since $H'$ and $H$ are sylow $3$-subgroups they are conjugate subgroups by Sylow's Second Theorem, so $gH'g^{-1} = H$. Hence, since $H' \leq N(K)$ we have that $H = gH'g^{-1} \leq gN(K)g^{-1} = N(gKg^{-1})$. Another application of Sylow's Second Theorem tells us that $gKg^{-1}$ is a sylow $5$-subgroup, say $K'$, so that $H \leq N(K')$. By the corollary that I mentioned above, this means that $HK'$ is a well-defined subgroup. Moreover, by order considerations this subgroup has order 15 which means it is abelian. Therefore, $HK' \leq N(H)$ as so $15 \mid |N(H)|$ and consequently $5 \mid |N(H)|$.