Prove that $G = \{e_G, a, a^2 , a^3 , . . . , a^{p−1} , b, ab, a^2 b, . . . , a^{p−1 }b\}.$

Question

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.

My guess is that you meant $G={e_G,a,a^2,a^3,\ldots,a^{p-1},b,ab,\ldots,a^{p-1}b}$. — José Carlos Santos, May 02 '20 at 07:54
I was talking about the title...but yeh I didn't see the body was not fixed Anything else you see? — PhysicsBish, May 02 '20 at 08:09

score 2 · Accepted Answer · answered May 02 '20 at 08:02

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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…

answered May 02 '20 at 08:02

José Carlos Santos

427,504

I still don't quit understand this to be honest. Can you explain more if possible. – PhysicsBish May 02 '20 at 08:11
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@PhysicsBish You can compare with many duplicates here, see for example this post, or this post or this post etc. – Dietrich Burde May 02 '20 at 08:18
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Can we have, for instance, $a^3=ab$ (when $p>3$)? No, because this is the same thing as asserting that $a^3b^0=a^1b^1$, we neither have $p\mid3-1$ nor $2\mid0-1$. – José Carlos Santos May 02 '20 at 08:38

Prove that $G = \{e_G, a, a^2 , a^3 , . . . , a^{p−1} , b, ab, a^2 b, . . . , a^{p−1 }b\}.$

1 Answers1