Prove that $G = \{e_G, a, a^2 , a^3 , . . . , a^{p−1} , b, ab, a^2 b, . . . , a^{p−1 }b\}.$ Let $G$ be a group of order $2p$ where $p$ is a positive odd prime.
By the $1^\text{st}$ Sylow Theorem, there exist $a, b \in G$ such that $|a| = p$ and $|b| = 2.$ Prove that $G = \{e_G, a, a^2 , a^3 , . . . , a^{p−1} , b, ab, a^2 b, . . . , a^{p−1 }b\}.$
There was a hint given: See Question 5 and assume $a^ib^j = a^kb^\ell$ with $0 \leq i, k \leq p − 1$ and $0 \leq j,\ell \leq 1$
and show $i = k$ and $j = \ell.$
And Question $5$ asks us: Let $G$ be a group and let $a, b \in G$ such that $|a| = n$ and $|b| = m.$ Suppose $⟨a⟩ \bigcap ⟨b⟩ = ⟨e_G⟩.$
Prove that $a^ib^j = a^kb^\ell$ if and only if $n$ divides $i − k$ and $m$ divides $\ell − j.$
I am just having trouble understanding how to approach this question. I have tried to give as much information as I could.

Since $a^ib^j=a^kb^l$ if and only if $p\mid i-k$ and $2\mid j-l$, the elements of $G$ $$e_G,a,a^2,a^3,\ldots,a^{p-1},b,ab,\ldots,a^{p-1}b$$are $2p$ distinct elements of $G$. And since $G$ has $2p$ elements…
