Identifying A $2p$ Group

Relevant wiki: Group Theory

We first would like to show that $G$ contains an element of order $p$ . This is simple enough if you are familiar with Cauchy's Theorem, but we do not actually need this. Suppose there is no element of order $p$ . Then, using Lagrange's Theorem , we know that the order of all non-identity elements must be 2. Take any two distinct non-identity elements $x,y \in G$ , and we have $xyx^{-1}y^{-1} = xyxy = (xy)(xy) = e_G$ , since $xy$ must be an element of order 2. Thus, equating $xyx^{-1}y^{-1} = e_G$ gives us $xy = yx$ . Since the identity commutes with all elements, this implies that $G$ is in fact abelian, contrary to the problem's assumption. There must exist an element of order $p$ , which we will call $r$ .

Immediately, we can rule out $\mathbb{Z}_p \times \mathbb{Z}_2$ and $\mathbb{Z}_2^p$ since these groups are both abelian; their group operation is addition, which is commutative. We realize that the group $P_y$ is cyclic (generated by $r$ , a rotation by $\pi / p$ about an axis perpendicular to the base and passing through the vertex) and therefore abelian, since $r^mr^n = r^{m + n} = r^{n + m} = r^nr^m$ . Thus, we rule out $P_y$ . $G$ does contain an element of the form $xyx^{-1}y^{-1} = e_G$ if both $x$ and $y$ are powers of $r$ .

We show that $G$ must be isomorphic to $P_r$ . Since $G$ has an element of order $p$ , $G$ must have a subgroup generated by $p$ with order $p$ : $G_r = \{e_G,r,r^2, \dots, r^{p-1}\}$ . Take an element $x \in G \setminus G_r$ . Since $x \notin G_r$ , and elements of groups have unique inverses, we have $x^{-1} \notin G_r$ . That is, $x^{-1}$ is a member of the $p$ elements of $G$ not in $G_r$ . Notice that since $p$ is an odd prime, one element of $G \setminus G_r$ must be its own inverse (we match pairs of inverse elements in $G \setminus G_r$ , but eventually we will have only 1 element unmatched). We call this element $y$ , and since $y = y^{-1}$ implies $y^2 = e_G$ , we see that $y$ has order 2. Since $y \notin G_r$ , the element $r^my$ must be in $G$ due to closure of a group, but it cannot be in $G_r$ since $r^my$ is not a power of $r$ . Therefore, all elements of the form $r^my$ are outside $G_r$ , and the elements of the subset $G \setminus G_r$ are $\{y,ry, \dots, r^{p-1}y\}$ . Each of these elements must have order 2, since if they had order $p$ , we would generate much more than $2p$ elements using $r$ and $ry$ . Therefore, $(r^my)(r^my) = e_G$ , or $r^my = yr^{-m}$ .

The group $P_r$ consists of two kinds of rotations: A rotation by $2\pi / p$ about an axis perpendicular to the base passing through the prism's center, and a rotation by $\pi$ about an axis parallel to the base passing through the prism's center which passes through an edge and the centroid of the face opposing the edge. The first rotation $t$ clearly has order $p$ , while the second rotation $s$ has order 2. The group is generated by these two rotations, giving us the elements $P_r = \{e_{P_r}, t, t^2, \dots, t^{p-1}, s, ts, t^2s, \dots, t^{p-1}s\}$ , and we can geometrically verify that the elements of the form $t^ms$ have order 2. We can also geometrically verify that $t$ and $s$ do not commute, and we have the obvious mapping $\phi : G \rightarrow P_r$ given by $\begin{aligned} & \phi : r \mapsto t \\ & \phi : y \mapsto s \end{aligned}$ which gives us an isomorphism from $G$ to $P_r$ , so $G \cong P_r$ .