$\textbf{Question.}$ Let $G$ a group of order $2010$ and suppose that $G$ has an abelian normal subgroup $N$ with order $6$.
a) Show that all elements of $G$ with order $2$ and $3$ are in $N$.
b) How many elements in $G$ has order $2$? How many elements in $G$ has order $3$?
c) Show that $N \leq Z(G)$.
I would like to know if what I did it's correct and how to proceed where I'm stuck. This is what I thought:
a) Let $\pi: G \longrightarrow G/N$ be the canonic projection. I know that $\pi$ is a homomorphism and $o(\pi(g))$ divides $o(g)$ for each $g \in G$, so if $o(g) = 2$, then $o(\pi(g)) = 1$ or $o(\pi(g)) = 2$, therefore, $g \in N$ or $g^2 \in N$. If $g \in N$, then I have what I wanted, otherwise, I have $g^2 \in N$ and I observed that $g^2 N \in \langle gN \rangle \bigcap N$. It seems that if $\langle gN \rangle \bigcap N \neq \emptyset$ (which is the case), then $gN = N$, because $G/N$ is a partition of $G$, but is it correct? I'm stuck here because I'm not sure if it's correct. If $o(g) = 3$, then I would like to apply the previous idea, but it depends if the affirmation that I did previously is correct.
b) By, I know that $\langle x \rangle \leq N$ for each $x \in G$ with $o(x) = 2$. Furthermore, $\langle x \rangle \trianglelefteq N$, because $N$ is abelian, then the element $x \in G$ with $o(g) = 2$ is unique to less than conjugation. By a similar argument, I concluded that $x \in G$ with $o(g) = 3$ is unique to less than conjugation.
Just an observation here: I thought that I needed use Sylows' theorem to do this item and use the fact that $|N| = 6$ in order to conclude that $\langle x \rangle \trianglelefteq N$, but it isn't really necessary, is it?
c) Since $N$ and $Z(G)$ are subgroups of $G$, it's sufficient show that $N \subset Z(G)$. Let be $g \in G$, $n \in N$ and observe that $[g,n] \in [G,G] \cap N$, because $gng^{-1} \in N$, then $[g,n] = (gng^{-1})n^{-1} \in N$, therefore $[G,G] \cap N \neq \emptyset$. I know that $[G,G] \cap N \trianglelefteq N$. By Lagrange's theorem, $|[G,G] \cap N| = 1, 2, 3$ or $6$. If $|[G,G] \cap N| = 1$, then I have what I want. I'm stuck in how to proceed in order to conclude that $N \subset Z(G)$ for the other cases.
Best Answer
Here is a solution to part $a$. Note that $|G/N|=\frac{|G|}{|N|}=335$. Now assume $g\in G$ is an element of order $2$. Then $g^2=e\in N$ when $e$ is the identity element of $G$. So that way we get $g^2N=N$. And then:
$N=N^{168}=(g^2N)^{168}=g^{336}N=gN(gN)^{335}=gN$
Where we use the fact that $|G/N|=335$ and hence $(xN)^{335}=N$ for all $xN\in G/N$. So we got $gN=N$ and that means $g\in N$. Now use the same idea for elements of order $3$.
Part $b$: I don't really understand your explanation, and anyway the answer is wrong. I'll give a hint: $\mathbb{Z_6}$ and $S_3$ are all the groups of order $6$ up to isomorphism. Your group $N$ is abelian, so it is $\mathbb{Z_6}$. Now it should be easy to count the elements of order $2$ and $3$. There is one element of order $2$ and two elements of order $3$.
Now to part $c$. I'll leave to you as an exercise to prove the following lemma:if $P$ is a $p$-group, $X$ is a finite set and $P$ acts of $X$ then the number of elements of $X$ with orbit of size $1$ is congruent to $|X|$ mod $p$. Now we'll use it.
Let $P$ be a $3$-Sylow subgroup of $G$. As we already know there are only $2$ elements of order $3$ we also know this is the only $3$-Sylow subgroup and hence it is normal in $G$. Now let $Q$ be a $67$-Sylow subgroup of $G$ and define an action of $Q$ on $P$ by $g.x=gxg^{-1}$. It is an action because $P$ is normal in $G$. Now let $X$ be the set of elements of $P$ with orbits of size $1$. These are the elements of $P$ that commute will all the elements of $Q$. You can check it is a subgroup of $P$, so its order must be $1$ or $3$. But $Q$ is a $67$-group and hence by the lemma $|X|$ is congruent to $|P|=3$ mod $67$. So $|X|$ can't be $1$, so it must be $3$, which means $X=P$. Every element of $P$ commutes with every element of $Q$ and that means $Q\leq C_G(P)$. By Lagrange's theorem we get $67$ divides $|C_G(P)|$.
Now do exactly the same process with a $5$-Sylow subgroup and conclude that $5$ divides $|C_G(P)|$. Also, $P$ is abelian and hence $P\leq C_G(P)$, so also $3$ divides $|C_G(P)|$. Finally, as there is only one element $g\in G$ of order $2$ we can conclude that this element is in the center of $G$ and hence $\langle g \rangle\leq C_G(P)$ so also $2$ divides $|C_G(P)|$. So as $2,3,5,67$ all divide $|C_G(P)|$ we get that $2010$ divides it too, so $C_G(P)=G$. That means $P$ is in the center of $G$.
Alright, so now look at the group $Z(G)\cap N\leq N$. The identity is there, the only element of order $2$ is there and as we showed both elements of order $3$ are there. So this group contains more than half of the elements of $N$ and by Lagrange's theorem we get that $Z(G)\cap N=N$. And that implies $N\leq Z(G)$.