Show that $N=\{g_1^3 \times \cdots \times {g_r}^3 \mid g_i \in G, r \geq 0\}$ where $g_i \in G$ and $r \ge 0$ is normal subgroup

Question

$N=\{g_1^3 \times \cdots \times {g_r}^3 \mid g_i \in G, r \geq 0\}$ where $g_i \in G$ and $r \ge 0$.

When $r=0$, we are to assume the product is the identity. All I am given is that $G$ is a finite group. I tried the following:

(I) I tried to find $N$ as a kernel of a homomorphism but to no avail.

(II) I tried showing $gng^{-1}$ is an element of $N$ but I'm not sure that we can assume $g$ can be written as a cube of another element. I tried doing induction on the order of $G$ but that doesn't help either.

Any hints on how to proceed?

Thank you

Answer 1

As given by the hint from @GerryMyerson, we will show that for any $g \in G$ and any $n \in N$, $gng^{-1} \in N$

Noting that $gg_1^{3}g^{-1} = (gg_1g^{-1})^3$ and that $gg_1^3g_2^3g^{-1} = (gg_1^3g^{-1})(gg_2^3g^{-1})$ we have that for $n = g_1^3 … g_r^3 \in N$ and $g \in G$,

$gng^{-1} = g(g_1^3 … g_r^3)g^{-1} = (gg_1^3g^{-1})…(gg_r^3g^{-1}) = (gg_1g^{-1})^3 … (gg_rg^{-1})^3 \in N$

Thus, N is indeed a normal subgroup.