Let $G$ be a group of order $30$. Show that if $G$ is nonabelian, there are more than one $2$-subgroups of Sylow.

Question

Let $G$ be a group of order $30$. Show that if $G$ is nonabelian, there are more than one $2$-subgroups of Sylow.

Suppose there is only one $2$-subgroup of Sylow and show that $G$ is abelian.

It is not clear how to use the hypothesis.

It has already been shown that there is a group $H$ of $G$ such that $H$ has order $15$, is cyclic and normal.

Let $x \in H$ such that $\langle x \rangle = H$.

Let $K$ the unique $2$-subgroup of Sylow.

Let, $y \in K$ a generator of $K$.

Since $H$ is normal in $G$, $yxy^{-1} = x^i$, for some $i \in \{1, 2, \cdots, 15\}$. For $y^2 = 1$, it follows that $$x = y^2 x y^{-2} = y (y x y^{-1})y^{-1} = y x^i y^{-1}$$ taking the power $i$, $$x^i = (y x^i y^{-1})^i = (y x^i y^{-1})\cdots (y x^i y^{-1}) = y x^{i^2} y^{-1}$$ and conjugating by $y$, $$x = y x^i y^{-1} = y^2 x^{i^2} y^{-2} = x^{i^2}$$ Therefore, $i^2 \equiv 1 \mod 15$. By exhaustion, $i \in \{1,4,11,14\}$.

If $i = 1$, $xy=yx$, but $HK = G$. Thus, $G$ is abelian.

Any ideas for the other cases?

Answer 1

If you have already prove that G has a cyclic subgroup of order 15 ,say $H$ then you are almost done. Clearly, $H$ is normal in $G$ ,since index of $H$ in $G$ is 2.

Now, if G has exactly one sylow-2-subgroup,say $K$,By sylow-2nd theorem, $K$ is normal in $G$ By calculating order of $HK$,we can say

$G=HK$

$H\cap K ={e}$

$G$ is isomorphic to $H×K$ That implies G is abelian.

Answer 2

Suppose there is only one, hence normal, Sylow $2$-subgroup, say $N$. Since normal subgroups are union of conjugacy classes, $N$ is central. Moreover, $G/N$ is cyclic (as $|G/N|=3\cdot5$ and $3\nmid 5-1$). But then $G$ is abelian. Contradiction.