Sylow and products of groups

Question

Consider $G$ a group of cardinal 105. There is in particular unique $5$- and $7$-Sylows. Let $P$ be a $3$-Sylow of $G$, and $N$ be the $7$-Sylow of $G$. Is that clear that $PN$ is a subgroup of $G$?

Is that a property particular to this case, or do we already have a subgroup when we multiply by a Sylow prime to the order of the other group?

Answer 1

It is a standard fact that if $H$ is a subgroup in any group $G$, and $N$ a normal subgroup of $G$, then $HN$ is a subgroup.

Answer 2

In general if $N$ is a normal subgroup of $G$ and $P$ is a subgroup then $PN$ is a subgroup of $G$. let's see why:

$\mathbf {Proposition}:$

Let $H,K \le G$ then $HK=KH \iff HK \le G$.

$\mathbf {Proof}:$

$\implies$

suppose $HK=KH$ then we have to show the following are true:

$e \in HK: e \in H$ (as $H$ is a subgroup) and $ e \in K$ (as $K$ is a subgroup), so $e=e \cdot e \in HK$.

$HK$ is closed under $G$'s multiplication: let $h_1k_1,h_2k_2 \in HK$.

Notice that since $HK=KH$ we have that there exists $hk \in HK$ such that $k_1h_2 = hk$.

So we have: $$h_1k_1h_2k_2 = h_1hkk_2 \in HK$$

$HK$ is closed under inverses: let $hk \in HK$, so $\left(hk\right)^{-1}=k^{-1}h^{-1}$.

Again, due to the fact $HK=KH$ we get that there exists $\bar{h}\bar{k} \in HK$ such that $k^{-1}h^{-1}=\bar{h}\bar{k} \in HK$.

$\impliedby$

suppose $HK \le G$ and we would like to show that $HK=KH$:

let $kh \in KH$. Notice that $k \in HK$ as $e \in H$ so $k = e \cdot k \in HK$, and the same goes for $h: e \in K$ so $h = h \cdot e \in HK$.

As $HK$ is a subgroup it is closed under $G$'s multiplication so $kh \in HK$. So we have $KH \subseteq HK$.

For the other direction, let $hk \in HK$, so there exists $\bar{h}\bar{k} \in HK$ such that $hk = \left(\bar{h}\bar{k}\right)^{-1} = \bar{k}^{-1}\bar{h}^{-1} \in KH$.

So we get $HK \subseteq KH$, which implies $HK=KH$.

$\mathbf {Proposition}:$

If $H \le G$ and $K \lhd G$ then $HK \le G$.

$\mathbf {Proof}:$

As we have the above proposition it is enough to show that $HK=KH$.

Because $K$ is a normal subgroup of G we have that for every $g \in G$ then $gK=Kg$.

So for every $h \in H$ we have $hK=Kh$, and now we get:

$$ hk \in HK \implies hk \in hK \implies \exists \bar{k} \in K : hk = \bar{k}h \in KH $$

Similarly we get the same for every $kh \in KH$ so we may conclude that $HK=KH$ which is equivalent to $HK$ being a subgroup of $G$, so we are finished.

I hope this helps.