Let $G$ be the group $S_4\times S_3$ . Prove or disprove the following:
I've got $|S_4\times S_3|=144$ and the group as not simple.
While applying Sylow's theorem I've got possible values of no. of Sylow $2 $subgroups as $9, 3, 1$ and no. of Sylow $3$ subgroups as $16, 4, 1. $ Then which value I've to exclude?
On
There are 3 subgroups of order 8 in $S_{4}$, let's call them $H_{1}$, $H_{2}$ and $H_{3}$. Let $\sigma$ be an element of order 2 in $S_{3}$. $H_{i} \times <\sigma>$ are subgroups of order 16, hence you have 3 of them at least, what can you conclude? Now as for 3 Sylows, you have a unique subgroup of order 3 in $S_{3}$ but you have, (I think), 4 subgroups of order 3 in $S_{4}$, what can you then conclude? (The idea here, was to explicitely construct p sylows, knowing the structure of the groups appearing in the direct product)
On
$\newcommand{\Size}[1]{\lvert #1 \rvert}$This seems to me one of those cases where working in a more general setting helps.
Suppose $H, K$ are two finite groups, and let $G = H \times K$. Since $\Size{G} = \Size{H} \cdot \Size{K}$, it is clear that the order of a Sylow $p$-subgroup of $G$ is the product of the orders of a Sylow $p$-subgroup of $H$ and one of $K$. Moreover, the product of a Sylow $p$-subgroup $H_{p}$ of $H$ and one $K_{p}$ of $K$ will yield a Sylow $p$-subgroup of $G$; this is because $H$ and $K$ commute elementwise, and thus $H_{p} K_{p}$ is a subgroup of $G$, of the right order.
Furthermore, $H_{p} K_{p}$ will be normal in $G$ if and only if $H_{p}$ is normal in $H$ and $K_{p}$ is normal in $K$ - just note first that if $h \in H$ and $k \in K$, you have $$ (H_{p} K_{p})^{h k} = H_{p}^{h} K_{p}^{k}. $$ And then, if $U \le H, V \le K$, then $UV$ uniquely determines $U$ and $V$ as $U = UV \cap H$ and $V = UV \cap K$.
Otherwise put (see Andreas' answer): $$Syl_p(G_1) \times Syl_p(G_2)=Syl_p(G_1 \times G_2).$$