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Let $G$ be a non-abelian group of order $2p$ for a prime $p>2$.

As a small part of an exercise I am trying to show that if we take $A, B \leq G$ s.t $|A|=p, |B|=2$ (existence guranteed by Sylow) then $AB=G$. but I'm having trouble showing that.

I thought about looking at $B \curvearrowright G/A$ by left multiplication, if we could show it is transitive that is enough I think.

Shaun
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paxtibimarce
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    $A$ is a normal subgroup, because it is of index $2$. So $G=A\ltimes B$ and hence $G=AB$. – Dietrich Burde Jun 14 '21 at 18:53
  • what I am trying to show is that G=A⋉B... don't you need G=AB first to claim it is true? – paxtibimarce Jun 14 '21 at 18:55
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    It depends on what you know. Strictly speaking it is a duplicate, since the only such group is $D_{p}$, which is a semidirect product. – Dietrich Burde Jun 14 '21 at 18:57
  • Why did you close my question? It is different... – paxtibimarce Jun 14 '21 at 19:00
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    It is exactly what you have asked for, namely why is it a semidirect product, i.e., $G=AB$ (and trivial intersection). So the answer is, because $G=D_p=C_p\ltimes C_2$ and hence $G=AB$. See also Shaun's answer below. In fact, it is your exercise. – Dietrich Burde Jun 14 '21 at 19:01
  • @DietrichBurde This is what I am trying to prove, and I'm looking to show specifically how G=AB without the entire theorem in the first place... So I suggest you don't close my question. – paxtibimarce Jun 14 '21 at 19:04
  • $G=AB$ basically is exactly that it is a semidirect product. It is really the same. – Dietrich Burde Jun 14 '21 at 19:05

2 Answers2

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Since $A$ has index $2$ in $G$, it must be normal (you can also show this with the sylow theorems, since $A$ is a sylow $p$-subgroup). But since $A$ is normal and $B$ is a subgroup, we know $AB \leq G$ (by one of the isomorphism theorems).

But of course, $|AB| = |G|$, so $AB$ must actually equal all of $G$!

I hope this helps ^_^

HallaSurvivor
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  • Thanks, but why $|AB|=|G|$? I don't think it is trivial since it should rely on $G$ being non-abelian. – paxtibimarce Jun 14 '21 at 18:58
  • We know that $|AB| = \frac{|A| \ |B|}{|A \cap B|}$, but since $A$ is a $p$-sylow subgroup it cannot contain any elements of order $2$, whereas $B$ contains exactly one element of order $2$, so their intersection is ${e}$, and $|AB| = |A| \ |B|$. – HallaSurvivor Jun 14 '21 at 19:01
  • Thanks again, but I'm a little confused since this should only be true if $G$ is non-abelian and you didn't use it. Am I missing something? – paxtibimarce Jun 14 '21 at 19:12
  • The claim is true even when $G$ is abelian. We just get a direct product $A \times B$ instead of a semidirect product $A \ltimes B$. This is actually still fine, since a direct product is just a very simple semidirect product. – HallaSurvivor Jun 14 '21 at 19:14
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Use the following

Theorem: Let $G$ be a group of order $2p$, where $p$ is a prime greater than $2$. Then $G$ is isomorphic to $\Bbb Z_{2p}$ or $D_p$.

For a proof, see Theorem 7.3 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)".

Shaun
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