If $|G|=2n$, $|a|=n$, $|t|=2$ and $tat=a^{-1}$ then $t\not \in \langle a\rangle$

Question

Let $G$ be a group of order $2n$ and $a,t\in G$ s.t $|a|=n, |t|=2$ and $tat=a^{-1}$. define $N=\langle a\rangle$.

I need to show that $|G:N|=2$ and $G=N\cup tN$.

to show that $|G:N|=2$ is trivial using Lagrange's Theorem but for showing that $G=N\cup tN$,$~~~$ I need to show that $t\not \in N$. how can I prove this?

Answer 1

This is not true if $n = 2, G = \Bbb Z_2\times \Bbb Z_2$, and $a = t$ is any element apart form the identity element, or if $G = \Bbb Z_4$ and $a = t = 2$. Also, if $G = \Bbb Z_2$ and $a = t = 1$, it is false that $G = N \cup tN$. Otherwise it's true.

Assume that $a \neq e$ and that $t \in \langle a\rangle$. That means that there is a $k$ such that $t = a^k$ (we set $0< k < n$). But then $tat = a^kaa^k = a^{2k+1} = a^{-1} = a^{n-1} = a^{2n-1}$. This implies that $2k = n-2$ or $2n-2$. But this gives either $$a^n = e = t^2 = a^{2k} = a^{n-2} = a^{-2}$$ or $$ a^n = e = t^2 = a^{2k} = a^{2n-2} = a^{-2} $$ which means that the order of $a$, and therefore $n$, is $2$ or $1$. The counterexamples in the first paragraph are therefore the only ones.

Answer 2

Since $N$ is cyclic, it is Abelian, so if $t\in N$, then $a^{-1}=tat=att=a$, which would imply that the order of $a$ divides $2$. Since $t\in\langle a\rangle$, we must have $|a|=2$. In that case, the statement is false for $t=a=2$ in $\Bbb Z_4$, or for $t=a$ in $V$, the Klein four group.