Let the $G$ is a group $s.t.$ its order is $150$
And Uniquely exists subgroup of the $G$ $s.t.$ its order is $6$
Show the existence of the subgroup whose order is $30$
When I solve this question, I don't have a confidence that my solution is right or not. Any advice or comments would be appreciated.
My attempt) $150 = 2\cdot 3 \cdot 5^2$
By the sylow thm, There are sylow 2, 3 subgroups in $G$ respectively.
Let the Sylow 2 group $P_2$ and sylow 3 group $P_3$
Since $H = P_2 \times P_3 $ and $H$ is a unique, Hence Each $P_2$ and $P_3$ are unique. (I.e. $P_2 \unlhd G, P_3 \unlhd G$)
Hence $H \unlhd G $ $s.t. \vert H \vert = 6$
Plus there are the other sylow 5 subgroup $P_5$ $s.t.\vert P_5 \vert = 25$
Surely By cauchy thm, $\exists g \in P_5$ $ s.t.$ $\vert g \vert = 5$
So, $\langle g \rangle \leq P_5$ and $\vert \langle g \rangle \vert = 5 $
Therefore $\langle g \rangle$$H$ is a subgroup of the $G$, and its order is $30$.
(1) $H$ is unique subgroup of order $6$, so it is normal.
(2) There exists subgroup $\langle g\rangle$ of order $5$. So product $\langle g\rangle H$ is subgroup, and its order is $30$.
The argument's of Sylow $2$,$3$-subgroups is unnecessary. Even instead of Sylow-$5$ subgroup, we need subgroup of order $5$ only; so modulo "Sylow" the rest of arguments give the proof.