# If $G$ is a non-abelian group of order 10, prove that $G$ has five elements of order 2.

I'm trying to prove this statement:

If $G$ is a non-abelian group of order $10$, prove that $G$ has five elements of order $2$.

I know that if $a\in G$ such that $a\neq e$, then as a consequence of Lagrange's theorem $|a|\in \{2,5,10\}$. The order of $a$ cannot equal $10$, since then $G$ would be cyclic, and thus abelian which is a contradiction. Now this means that $|a|=2$ or $|a|=5$. I know from this question that $G$ has a subgroup of order $5$. This subgroup $H$ has prime order, so it is cylic, and all of its non-identity elements have order $5$. Now I need to show that the elements not in $H$ have order $2$. This is where I'm stuck.

I've tried assuming that an element $b \notin H$ has order $5$, in order to derive a contradiction, but to no avail.

I also know from a previous exercise that if $G$ has order $10$, then it has at least one subgroup of order $2$, so I tried to assume toward a contradiction that $G$ has two subgroups of order $5$, and one subgroup of order $2$. I was trying to show that this would make $G$ abelian, but I couldn't.

Any ideas?

• Have you covered Sylow theory? It will immediately follow from Sylow theorems that there is exactly one subgroup of order five. If you haven't done Sylow theory, then you need a bit of tinkering (or conclude normality of the subgroups by some means to follow the route hinted at by DonAntonio). This is a small enough case that can be done without using much theory at all. – Jyrki Lahtonen Jul 24 '16 at 19:28
• Thanks, I have not covered sylow theory yet. – M47145 Jul 24 '16 at 19:32
• Ok. Are you familiar with conjugacy classes? Or with the fact that elements in a conjugacy class are all of the same order? – Jyrki Lahtonen Jul 24 '16 at 19:33
• No, so far the book has covered, Lagrange's theorem, Cayley's theorem, congruence classes, cosets, isomorphisms and homomorphisms. Normal subgroups, conjugation, quotient groups, Sylow theory etc are still to come. – M47145 Jul 24 '16 at 19:39
• Ok. I try to cook up an argument using available technology only. – Jyrki Lahtonen Jul 24 '16 at 19:40

There are only 2 groups of order 10, namely the cyclic group and the dihedral group of symmetries of a regular pentagon. The reflections in the dihedral group give you the five desired elements of order $2$.

Now, to prove that there are only 2 groups of order 10, let $a,b$ be elements of orders $2,5$ respectively. Consider the elements $1$, $b$, $b^2$, $b^3$, $b^4$, $a$, $ab$, $ab^2$, $ab^3$, and $ab^4$, and notice that they are all distinct. To determine the group, it suffices to determine what $ba$ is. This is the same as determining what $a^{-1}ba$ is. By nonabelianness, we know that $a^{-1}ba \neq b$, so merely check against all other elements of the group...

• In a sense that is what I'm trying to prove, so I'm not sure how this helps. I can't assume that there are only two groups of order $10$. – M47145 Jul 24 '16 at 19:30
• I have now edited my response to include a method for approaching your most recent question. – Ashvin Swaminathan Jul 24 '16 at 19:37
• Quick question, why do we know that $a^{-1}ba \neq b$? Nonabelian means that there is at least one pair of elements that do not commute. How do I know that those elements are $a$ and $b$? – M47145 Jul 24 '16 at 19:55
• Good question! Notice that if $ab = ba$, then every pair of elements in the group must actually commute! – Ashvin Swaminathan Jul 24 '16 at 19:56
• Hmm. How do you conclude in your second solution that $a^{-1}ba$ is a power of $b$? I think the OP is trying to prove that the elements $ab^i, i=1,2,3,4$ all have order two (not order five), and you seem to be assuming that they are not of order five... – Jyrki Lahtonen Jul 24 '16 at 20:03

Hint:

It has one single subgroup $\;P\;$ of order two and one single subgroup of order five $\;Q\;$ iff $\;P,Q\lhd G\;$ , but then $\;G=PQ=\ldots\;$

Use *Sylow's theorems:

The number of subgroups of order $5$ is congruent to $1$ modulo $5$, and a divisor of $\frac{10}5=2$. As there can't be $6$ since $G$ would have at least $25$ elements, there's only one subgroup of order $5$, say $H$, which is therefore a normal subgroup of $G$, generated by any of its elements $g$ distinct from $e$.

Let $a$ be an element of order $2$; its conjugate $gag^{-1}$ also has order $2$, and it is distinct from $a$, since this would mean $ga=ag$. The same would be true for any power of $g$, so that $G$ would be abelian.

Thus there are $5$ elements of order $2$.

The OP had worked out that the group $G$ has a subgroup $K$ of order $5$ (generated by $b$) and subgroup $H$ of order two (generated by an element $a$), and that all the non-identity elements must have order five or two.

Because $K\cap H=\{1_G\}$ the subsets $H$ and $K$ contain six elements between them, so four others remain. Our goal is to show that all those four elements must have order two. The contrapositive assumption is that at least one of them has order five, so let's assume that there is an element $c\in G, c\notin K$ such that $c$ has order five. The cyclic subgroup $K'$ generated by $c$ has five elements. Because $K\cap K'$ is not all of $K$ it must be trivial. Because $K'$ has an odd number of elements, it cannot contain $a$ either. So $K'$ must contain all the four elements outside of $K\cup H$.

Summary of the situation: $$G=\{1,a,b,b^2,b^3,b^4,c,c^2,c^3,c^4\}$$ with $a^2=1=b^5=c^5$. Here the elements have orders $1,2,5,5,5,5,5,5,5,5$ respectively.

Let us next consider the element $a'=bab^{-1}$. It should be one of the listed elements, but which? Let's try to figure out its order! Could we have $a'=1$? No! If $bab^{-1}=1$, then $$1=b^{-1}1b=b^{-1}(bab^{-1})b=(b^{-1}b)a(b^{-1}b)=1a1=a,$$ which is absurd. What about $a'^2$? Let's see! $$a'^2=(bab^{-1})^2=bab^{-1}bab^{-1}=ba(b^{-1}b)ab^{-1}=baab^{-1}=ba^2b^{-1}=b1b^{-1}=1.$$ So we can conclude that $a'$ has order two. But checking our list we see that we must have $a'=a$. That's good news, because now we can conclude that $$ba=bab^{-1}b=a'b=ab.$$ In other words $a$ and $b$ must then commute. Now let's consider the element $d=ab$. I want you to work out that

1. $d$ is not equal to either $1$ or $a$, and must therefore be of order five.
2. But $d^5=(ab)^5=ababababab=a^5b^5=???$ what is it? Why is this a contradiction?
3. Why does this prove your claim?
• Thanks for taking the time to write that! 1. If $d=1$ then $b=a^{-1}=a$, which is false. If $d=a$, then $b=1$, which is false. 2. $d^5=a$, which is false since $d^5= e$ and $a\neq e$. 3. This proves my claim since $d$ must be equal to $c,c^2,c^3$ or $c^4$. But in any of these cases $|d|=5$, which leads to the contradiction in 2. Thus $K'$ can't be a subgroup of order $5$. So no element in $K'$ can have order $5$. The only possibility is that all non-identity elements of $K'$ have order $2$. – M47145 Jul 24 '16 at 20:47
• Correct, @M47145. This is probably not the solution the author of your question intended. I don't remember how I might have solved this when beginning on group theory :-/ – Jyrki Lahtonen Jul 24 '16 at 21:12

There is a much simpler solution than using Sylow or $D_5$ groups. As OP said, since non-abelian G has order 10, and there must be a subgroup of order 5, then by Lagrange's Theorem there are two cases:

1. 5 subgroups of order 2, and 1 subgroup of order 5
2. 2 subgroups of order 5, and 1 subgroup of order 2

Assume the first case is true: We have elements a and b of order 5 in G. Then $a^5 = e = b^5$. Since $ab \neq a \Rightarrow b = e$, and $ab \neq b \Rightarrow a = e$, we must have ab in G where $(ab)^5 = e$, which means there are 3 subgroups of order 5, and we have our contradiction. Thus case 1 is true.

• By ”assume the first case”, you mean the second case. Also, do you mean $b\neq e$ and $a\neq e$? – M47145 Dec 12 '17 at 2:48