Show that a group of order $2010$ with an abelian normal subgroup of order $6$ is abelian

Question

Let $G$ be group of order $2010$ with $N$ an abelian normal subgroup of order $6$; show:

1. $N\le Z(G)$ where $Z(G)$ is the center of $G$
2. Show that there exist a unique $5$-Sylow subgroup of $G$
3. Conclude that $G$ is abelian

For the first question, note that there exists an element $x$ of $N$ with order $3$ and an element $y$ with order $2$; we can show that both $\left\langle x \right\rangle,\left\langle y \right\rangle$ are characteristic in $N$ (so normal in $G$); this implies that $N=\left\langle x \right\rangle \times \left\langle y \right\rangle$ and we conclude by observing that $x,y \in Z(G)$ (by an argument related to conjugacy classes).

But I'm stuck on the second question; how would you show that $P\in \text{Syl}_{5}(G)$ is unique?

For the third part, I think it's fairly straightforward as all Sylow subgroups $(2, 3, 5, 67)$ are normal, so $G$ is the direct product of such subgroups.

Answer 1

Here is an outline. Let $G$ be a group of order $2010$, and let $N$ be a normal subgroup of $G$ with $|N| = 6$.

You need to show that $G$ has a normal $5$-Sylow, so let $R \leq G$ be a Sylow $5$-subgroup of $G$. Prove that $RN$ is a subgroup of order $30$, and that $R$ is normal in $RN$.

Thus $N_G(R)$ has order divisible by $30$. Apply Sylow's theorem to conclude that $N_G(R) = G$ and $R$ is normal in $G$.

Answer 2

Slightly different arguments (since I do not see the solution of @testaccount clearly). Since $N \cong C_6$, we can apply the $N/C$ Theorem: $G/C_G(N)$ embeds isomorphically into Aut$(N) \cong C_2$. But $N \subseteq C_G(N)$ since $N$ is abelian, and $|G:N|=5 \cdot 67$, here $G/C_G(N)$ must be trivial, so $G=C_G(N)$, equivalent to $N \subseteq Z(G)$. This implies that the Sylow $2$-subgroups and Sylow $3$-subgroups are unique and so normal in $G$.

Since $|G/N|=5 \cdot 67$, and there is only one group of order $5 \cdot 67$ (apply Sylow Theory), we have $G/N \cong C_{335}$. If $R \in Syl_5(G)$, then $RN/R \in Syl_5(G/N)$, yielding $RN \unlhd G$. But $N$ is central so $R \unlhd RN$, even $R \text{ char } RN \unlhd G$, hence $R \unlhd G$.

Finally, it is obvious that there is only one Sylow $67$-subgroup, hence all Sylow subgroups are normal and cyclic of prime order, and so $G$ is cyclic.