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There are a few question on classification of groups of order $2p$ on MSE but I'd like to receive a feedback on this proof (and have a question about it at the end).

Let $G$ be a group of order $2p$. If there is an element of order $2p$ in $G$, then the group is cyclic. Suppose there is no such element. We claim that $G\simeq D_p=<a,b| a^p=1,b^2=1, baba=1>$.

By Cauchy's theorem, $G$ contains an element $a$ of order $p$. Then it contains its cyclic subgroup $H=\{1,a,\dots,a^{p-1}\}$. Let $b$ be an element outside of this subgroup. The $p$ elements $b, ba,\dots ba^{p-1}$ are pairwise different (if $ba^k=ba^m$ for $0\le k,m < p$, then $a^k=a^m$, which contradicts to the fact that $H$ contains pairwise different elements). Moreover for no $k$ and $m$ as above can the equality $a^k=ba^m$ hold (if it holds, then $b=a^ka^{-m}$, which contradicts that $b\notin H$). Thus $G=H\cup bH$ as a set.

We proceed to prove that $b^2=1$. It cannot be the case that $b^2\in H-\{1\}$ because if $b^2=a^r$ for $0 < r < p$ then $a=(a^r)^l=(b^2)^l\implies b=a^{-2l}$, which contradicts to the fact that $b\notin H$ (here we use that $H$ can be generated by any non-identity element because $p$ is prime; in particular, $H=<a^r>$ so that $a=(a^r)^l$ for $l$ an integer). We cannot have $b^2\in bH$ either because $b^2=ba^k$ yields $b\in H$. Thus $b^2=1$.

Also $(ba)^2=1$. Indeed, if $baba\in bH$, then similarly $b\in H$. If $baba\in H-\{1\}$, then $baba=a^r\implies bab=a^{r-1}\implies b^2=a^{r-2}$, and as above, $a=(a^{r-2})^l=(b^2)^l\implies b=a^{-2l}$.

So $G$ is generated by $a,b$ and the relations $a^p=b^2=baba=1$ hold in $G$.

Is it sufficient to conclude that $G$ is isomorphic to $D_p$? Or do I need to verify somehow that there are no relations in $G$ other than those implied by the ones above? And are there any other gaps or mistakes in my proof?

user557
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1 Answers1

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You already know that every element has the form $a^ib^j$ for appropriate exponents $i,j$. Furthermore, you can calculate any product $a^ib^ja^{i’}b^{j’}$ using the conjugation rule $bab=a^{-1}$.

Therefore, you have the complete multiplication table for the group. Since the dihedral group has that same multiplication table, your unknown group is indeed isomorphic to the dihedral one.

Alon Amit
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  • I guess to say that every element is of the form $a^ib^j$ I have to prove that $ba^s=a^{-s}b$ (which follows easily from $ba=a^{-1}b$). Then every element $ba^i$ from $bH$ can be written as $a^{-i}b$; the square of any such element can be shown to be $1$; the product of an element from $H$ by an element from $bH$ viz. $a^iba^j$ can also be written as $a^ia^{-j}b=a^{i-j}b$, and $ba^ja^i=ba^{i+j}=a^{-i-j}b$. Finally, $a^ib^ja^kb^m$, according to my calculations is $a^{i\pm k}b^{m+j}$ (using $ba^k=a^{-k}b$, not sure how to only use $bab=a^{-1}$). – user557 Aug 06 '18 at 18:08
  • To say that every element has the form $a^ib^j$ you just need to count. There are $p$ elements $a^i$ and $p$ more of the form $a^ib$. – Alon Amit Aug 06 '18 at 18:14
  • I’m not sure I follow the second question. Do you see how to compute the product of any two elements? – Alon Amit Aug 06 '18 at 18:15
  • For your first comment: but we don't know a priori that elements from $HbH$ (those of the form $a^iba^j$) lie in $G$ (i.e., that $G$ is closed under multiplication), that's why I mentioned the argument with transforming them to the form $a^{i-j}b$. So I don't see why just counting works. (And as I'm writing this comment, I'm realizing that we also need to prove that the inverse of all products of elements of $G$ also have the form $a^ib^j$, this also seems a hassle...) – user557 Aug 06 '18 at 18:23
  • For your second comment, I think I can see how to transform $a^ib^ja^kb^m$ to the form $a^rb^s$, but only if I use the relation $ba^q=a^{-q}b$, not just the relation $bab=a^{-1}$, which you mentioned. – user557 Aug 06 '18 at 18:25
  • I’m confused. $G$ is a group. What do you mean by “we don’t know that it’s closed under multiplication”? – Alon Amit Aug 06 '18 at 18:31
  • Actually now I'm confused too. We started by assuming $G$ is a group. Then we showed that as a set it has the form $H\cup bH$. Does it follow automatically (by our assumption) that $H\cup bH$ is a group? – user557 Aug 06 '18 at 18:38
  • Of course it is, as a set with the same binary operation. It is the same set and the same operation as $G$. I’m not sure what is the source of your confusion. – Alon Amit Aug 06 '18 at 18:42
  • Then why do we need to prove that the product of two elements of the form $a^ib^j$ is again of this form? If we know that $G$ is a group and every element in it is of the form $a^ib^j$, then any product will automatically have the form $a^ib^j$. (My understanding is that you are saying in the first paragraph of your answer that we need to prove this, but I may have misunderstood you). – user557 Aug 06 '18 at 18:52
  • We don’t need to prove that it has this form. We just need to prove that which element of this form is the product of two such elements is already well-defined by the properties of $G$ we have unraveled. – Alon Amit Aug 06 '18 at 18:54
  • I see, so in the first passage of your answer you're just saying that if we have two elements of the form $a^ib^j$, then the group relations enable us to specify concrete $k,l$ so that the product is equal to $a^ka^l$, not just conclude abstractly that the product is of the form $a^{i'}b^{j'}$. Further, $D_p$ has the same multiplication table because it has the same relations as those we used to compute the above mentioned product. But still, does it imply that $G$ doesn't have extra relations that are not present in $D_p$? – user557 Aug 06 '18 at 19:12
  • A group is fully determined by its multiplication table. You can always add more relations that are consequences of ones you’ve found, so identifying isomorphism by relations isn’t always obvious. Since they have the same multiplication table, it follows that corresponding generators satisfy the same relations. – Alon Amit Aug 06 '18 at 19:14