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In Gallian's Contemporary Abstract Algebra, 9th edition, on page 145, he is trying to prove the following: "Let $G$ be a group of order $2p$ , where $p$ is a prime greater than $2$. Then, G is isomorphic to $\mathbb{Z}_{2p}$ or $D_p$". He first proves that G must have an element of order $p$, which he names $a$. The following part of the proof is confusing to me:

Now let $b$ be any element not in $\langle a \rangle$, then by Lagrange's Theorem and the assumption that G does not have an element of order $2p$, we have that $|b|=2$ or p. Because $|\langle a \rangle \cap \langle b \rangle|$ divides $|\langle a \rangle|=p$ and $\langle a \rangle \neq \langle b \rangle$we have that $|\langle a \rangle \cap \langle b \rangle|=1$.

This part I am confused about. How does he conclude that? I can only think that $|\langle a \rangle \cap \langle b \rangle|$ is a subgroup, so its order must divide $2p$.

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    It is a subgroup of the group generated by $a$ so the order must divide $p$. – Mark Bennet Mar 01 '18 at 22:15
  • Why is that the case? How come $|\langle a \rangle \cap \langle b \rangle|$ is a subgroup of $\langle a \rangle$ –  Mar 01 '18 at 22:17
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    Because it is a group and it is contained in $\langle a \rangle$ because it is a subset (intersection). – Mark Bennet Mar 01 '18 at 22:27
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    https://math.stackexchange.com/questions/739237/let-g-be-a-group-of-order-2p-where-p-is-a-prime-greater-than-2-then/739240#739240 – Ivan Di Liberti Mar 01 '18 at 22:38
  • @IvanDiLiberti Thank you, I am aware of that question but homomorphism is not covered in the book yet, nor Cauchy's Theorem and the post of the question doesn't go over this detail. –  Mar 01 '18 at 22:47
  • @MarkBennet I see, thank you! –  Mar 01 '18 at 22:48

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More generally, if $H$ and $K$ are subgroups of $G$, then $H\cap K$ is a subgroup of $G$. Indeed, let $x,y \in H\cap K$ then $y^{-1} \in H\cap K$ since $y$ is both in $H$ and in $K$ which are subgroups. And thus $xy^{-1} \in H \cap K$ for the same reason.

So in this case, $\langle a \rangle \cap \langle b \rangle$ is a subgroup of $G$, but since it's a subset of $\langle a \rangle$ it's also a subgroup of $\langle a \rangle$, and so its order, which is not $p$, divides $|\langle a \rangle| = p$ which is prime, so it's $1$.

krirkrirk
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