# Number of elements in a Group such that $x^7=e$

Given that $$G$$ is a finite Group. Prove that number of elements in G such that $$x^7=e$$ where $$x \in G$$ is always Odd.

My attempt:

First possibility is a Trivial group since $$e^7=e$$. Trivial Group contains One element which is Odd number.

Now let $$x\ne e$$,then since $$x^7=e$$ and $$7$$ is prime we have $$ord(x)=7$$ and this implies $$ord(x^i)=7$$ $$\:$$ $$\forall 1\le i\le 6$$

So minimum possible non trivial group is $$G=\left\{e, x,x^2, x^3,x^4,x^5,x^6\right\}$$ Which is a Cyclic Group containing $$7$$ elements which is Odd.

Next possibility is another Cyclic group with $$13$$ elements which is Odd. So in general the Cardinality of the group is of the form $$6n+1$$.

But are all these cyclic groups only? Does it mean if $$ord(x)$$ is Finite, then the group is Cyclic?

• See this question and replace $3$ by $7$. Commented Apr 30, 2020 at 12:17
• If an element has order 7, its inverse does as well Commented Apr 30, 2020 at 12:18
• Re your last question, if $ord(x)$ is finite for all $x\in G$, then $G$ need not by cyclic. Trivially, $ord(x)$ is finite for every $x\in G$ already when $G$ is finite. Beyond that, $ord(x)$ is also finite for all $x$ in the infinite (and non-cyclic) group $\Bbb Q/\Bbb Z$, wheras most elements of the (cyclic!) group $\Bbb Z$ have infinite order. Commented Apr 30, 2020 at 12:20

In general, if $$p$$ is an odd prime and $$G$$ a finite group, then $$\#\{ g \in G\mid g^p=e\} \equiv 1 \bmod (p-1).$$

• But according to theorem $3.2$ in this link kconrad.math.uconn.edu/blurbs/grouptheory/order.pdf, If $x^n=e$ then the Group can contain only either $1$ element(Trivial) or $n$ distinct elements for any positive integer $n$. So if $n=7$ only two such groups are possible with $|G|=1,7$ Commented Apr 30, 2020 at 13:28
• No, we don't have $x^7=e$ for all elements in the group here. We only consider the set of elements which does satisfy it in a given finite group. Commented Apr 30, 2020 at 13:30
• But all integral powers of $x$ i.e., $x^i$ satisfy $x^7=e$, do you mean there might be some $y \in G$ such that $y^7 \ne e$ Commented Apr 30, 2020 at 13:35
• Yes, our question only asks about the cardinality of the set $S=\{x\in G\mid x^7=e\}$ in $G$. Why should be $S=G$? Commented Apr 30, 2020 at 13:38
• Thank you, now its clear, i misunderstood the problem Commented Apr 30, 2020 at 13:45

There's another way :

Think about it. If a group $$G$$ is cyclic and order of an element $$a$$ is $$7$$ then $$$$ is the only subgroup containing all the elements of order 7, which are $$\phi {7} = 6$$ in number and also for Identity $$e$$ $$e^7= e$$ too. So there's only odd number of such elements $$x$$ in $$G$$.

If a group is not a cyclic group, then if a be an element of order 7 then $$$$ contains 6 elements of order $$7$$. If this exhausts all order $$7$$ elements then we have $$6+1=7$$ i.e odd number of such elements $$x$$.

If there's another element $$b$$ of order 7 which doesn't belong to $$$$ then $$ \cap = \phi$$. Thus the $$6$$ elements of $$$$ i.e. $$\{b, b^2,....,b^6 \}$$ are distinct elements of order 7 and now the total number of order 7 is $$2\phi(7) = 12$$. If this process exhausts all the elements of order $$7$$ then we are done otherwise going with the same argument we have that the number of order 7 elements in this group is a multiple of $$\phi(7)$$ i.e. $$n\phi(7) = 6n$$ for some integer $$n$$.

Therefore the number of such $$x$$ is $$6n+1$$, which is odd.

So you can see that if $$|x|=7$$ then it's not necessary that the group must be cyclic.

• very good explanation...awesome Commented May 3, 2020 at 16:45