# how are group order and element order related?

Is there a relationship between group order and element order?

I know that there is a relationship between group order and subgroup order, which is that $[G:H] = \frac{|G|}{|H|}$ where $H$ is the subgroup of $G$ and $[G:H]$ is the index of $H$ in $G$. But is there a relationship between group order and the order of elements in the group?

For example, let the group $G$ be of order $7^{3}$. Does $G$ have an element of order $7$?

## 2 Answers

Note that by definition the order of an element is the order of the group generated by it, i.e we have that $|a| = |\langle a \rangle|$. Now obviously $\langle a \rangle \le G$, so we can easily correlate an order of an element with the order of the corresponding subgroup.

This proves that if $a \in G$, then $|a| \mid |G|$. On the other side by Sylow Theorems we have that if a prime power $p^n$ divides the order of $G$ then there's a subgroup of order $p^n$ in $G$. On the other side even if a composite number divides the order of the group $G$, the existence of a subgroup of that order can't be guaranteed.

• what does $|a| \mid |G|$ mean? – PiccolMan Oct 23 '16 at 21:08
• The order of the element $a$ divides the order of the element $G$. – Stefan4024 Oct 23 '16 at 21:09
• If a prime power $p^n$ divides $|G|$, that does not imply that there's an element of order $p^n$ in $G$. (Only a subgroup of that order!) – verret Oct 23 '16 at 23:24
• @verret comment helped to understand this answer. Thanks. Also, can anyone link me to the wiki section of Sylows theorem that talks about this. – PiccolMan Oct 24 '16 at 1:59

The order of any element must divide the group order if the group is finite. This follows from what you have written i.e, the Lagrange's theorem.

Your second question is a consequence of the Cauchy's theorem.