# 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.

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:

1. 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.
2. 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.
• In the abelian case: how do you know that it is not possible that there are no elements of order 2? Commented Feb 11, 2017 at 11:00
• @Student $g^p$ is such. Commented Feb 11, 2017 at 11:01
• Of course, thank you! Commented Feb 11, 2017 at 11:03
• One last question: it took me quite some work to see that in the second case $g^p \neq e$. Is there an easy way to see this? Commented Feb 11, 2017 at 15:15
• @Student Well, $g^{p+1}\in H$, because $g^2\in H$ and $p+1$ is even. But if $g^p=e$, then $g=g^{p+1}\in H$ - a contradiction. Commented Feb 11, 2017 at 17:50

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.

• Seems like only basic knowledge can be assumed? Commented Feb 11, 2017 at 10:43
• @pepa.dvorak: I'm sorry but I don't understand your question.
– Surb
Commented Feb 11, 2017 at 10:44
• He/She mentions "I only saw the definition of groups, subgroups, cosets and order of groups/subgroups and elements", so I understand it as "prove it using only basic facts about groups". Commented Feb 11, 2017 at 10:47
• Yes that's what i meant. This is a course i tool three years ago, so i do know more about group theory, but i'd like to solve this exercise with only the basics (since at that moment i only knew about those basics) Commented Feb 11, 2017 at 10:55
• @surb: thanks for your answer though, i had not thought about the sylow theorem, since i wanted to use the basics only. Commented Feb 11, 2017 at 10:56

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 $2$.

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.

• In te case where it is $\mathbb{Z}_{2p}$, shouldn't you have only one element of order 2? Any abelian group with at least two elements of order 2 has a subgroup of order 4 which is not possible by lagrange's theorem. Commented Feb 11, 2017 at 22:25
• Correct, I've comprised the neutral element as well (in my head I've beet working with the set $\left\{g\in G ; g^2=1\right\}$. Commented Feb 12, 2017 at 8:15