Number of elements of order 2 Suppose $G, \ast$ is a finite group or order $2p$, where $p$ is an odd prime. I was able to prove that there must be a subgroup of order $p$ denoted by $H$ and that for each element $g \in G$ we have that $g^2 \in H$. The third question I needed to solve regarding this exercise is the number of elements of order $2$ in the group $G$. I think I should be able to use what I found before. 
I also know (from a previous exercise) that each abelian group having at least two elements of order 2 has a subgroup of order 4, so for abelian groups, I have at most 1 element of order 2. I also know that the dihedral group of order $2p$ has $p$ elements of order 2. However, I have no idea on how to prove the number of elements more generaly... Any hints? (I only saw the definition of groups, subgroups, cosets and order of groups/subgroups and elements). Thank you in advance.
 A: Let $H\subset G$ be a group of $p$ elements. $H$ is both left and right coset and it is easy to see that then $G\backslash H$ [just in case: elements of $G$ that are not in $H$] is also both left and right coset. If you take $g\in G\backslash H$, then all the elements of the kind $gg_1$, where $g_1\in G$ are different. Because for $g_1\in H$ we already get all $p$ elements of $G\backslash H$ as $gg_1$, it follows that $g^2\in H$. Now there are two cases:


*

*For all $g\in G\backslash H$ we have $g^2=e$. Then there are $p$ elements of the order $2$, just like in dihedral group.

*For some $g\in G\backslash H$ we have $g^2\in H$ but $g^2\ne e$. Therefore, $g^2$ is of the order $p$, and $g$ - of the order $2p$ - abelian case.

A: All element of order $2$ are in a $2-$Sylow group. If $n$ is the number of $2-$Sylow group, then $n\mid p$ and $n\equiv 1\pmod 2$, therefore, $n\in\{1,p\}$. Therefore, there is either $1$ element of order 2 or $p$ element of order $2$, but you can't say more with your hypothesis.
A: A finite group of order $ 2p $, where $ p $ is an odd prime, is either cyclic of order $ 2p $, or the dihedral group $ D_p $. This follows easily from Sylow, but one can argue as follows without using any "advanced" results: First, we prove that $ G $ has a normal subgroup of order $ p $. Indeed, any subgroup of order $ p $ must be normal, since subgroups of index $ 2 $ are always normal (why?).
Now, it is easy to see that $ G $ is the internal semidirect product of a normal copy of $ C_p $ and $ C_2 $: it is isomorphic to $ C_p \rtimes C_2 $. Since $ \textrm{Aut}(C_p) \cong C_{p-1} $, there are two homomorphisms $ C_2 \to \textrm{Aut}(C_p) $, which give rise to two different semidirect products. One is the direct product $ C_2 \times C_p \cong C_{2p} $, and the other is the group whose presentation is 
$$ \langle x, y : x^p, y^2, yxyx \rangle $$
which is just the dihedral group $ D_p $. The dihedral group $ D_p $ has $ p $ elements of order $ 2 $, whereas $ C_{2p} $ has only one element of order $ 2 $.
A: I've given it a (slightly) different try:
If all elements are of order 2 then $G$ is abelian, hence isomorphic to $\mathbb{Z}_{2p}$, which is contradiction.
So there exists an element $g$ of order $\neq 2$.


*

*if $o(g)=2p$, you have $\mathbb{Z}_{2p}$ and hence $1$ element of order $2$.

*if $o(g)=p$, then the subgroup $H$ generated by $g$ is one coset and the other is $S=aH=Ha$ for some (any) $a \notin H$, both of them are of cardinality $p$. Now take $b \notin H$, hence $b = ag^k$ and notice that for such $b$ there exist a unique exponent $m$ such that $ag^m$ is the inverse of $b$, hence $$ag^kag^m=1 $$
which is equivalent to $ag^ka = g^{-m}$.
Now $$b^2 = ag^kag^k = ag^{-m}g^k = g^{k-m}.$$
Since the exponents are taken modulo $p$ and for each $0<n<p$ there exists an inverse, were $k-m \neq 0$  there would exist "square root" of $b^2 \in H$, hence $b\in H$, a contradiction. Therefore $k-m = 0$ and $b^2 = 1$, which shows that all $p$ elements of $S$ are of order two.
