Non-Cyclic Group of Order 55 has Order 5 Element and Order 11 Element

Example of Order of Group Element

Let $G$ be a non-cyclic group whose order is $55$.

Then $G$ has:

at least $1$ element of order $5$

at least $1$ element of order $11$.

Proof

By Order of Element Divides Order of Finite Group, all elements of $G$ have order in $\set {1, 5, 11, 55}$.

But as $G$ is non-cyclic, it can have no element of order $55$.

By Identity is Only Group Element of Order 1, $G$ has $54$ elements of order $5$ and $11$.

Let $m$ denote the number of subgroups of $G$ of order $5$.

Let $n$ denote the number of subgroups of $G$ of order $11$.

From Number of Order p Elements in Group with m Order p Subgroups, $G$ has:

$4 m$ elements of order $5$

$10 n$ elements of order $11$.

Then:

$4 m + 10 n = 54$

As neither $4$ nor $10$ divide $54$, it follows that neither $n$ nor $m$ are equal to $0$.

Hence the result.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $19 \ \text {(i)}$