Question

If G is a group of order 25, prove that either G is cyclic or else every non-identity element of G has order 5.

Short answer

We proved that if G is a group of order 25, then is either cyclic or every non-identity element of G has order 5.

Step-by-step solution

To determine the order of  x

We have the definition of a subgroup that,

Let $(G,*)$ be a group under binary operation, say $*$. A non-empty subset H of G is said to be a subgroup of G, if localid="1652329676024" $(H,*)$is itself a group.

We know the definition of a cyclic group.

A group G is called cyclic if there exists an elementlocalid="1652329799404" $a\in G$ such that every element of G can be written as $a^n$, for some $n\in Z$.

Let G be a group of order 25 i.e., $o\left(G\right)=25$.

We have to show that G is either cyclic or every non-identity element of G has order 5.

Suppose G is not cyclic.

Let $x\in G$be any element such that $x\ne e$.

Then by Lagrange’s Theorem,

$o\left(x\right)\left|o\right(G)$

$\Rightarrow o\left(x\right)|25$

Thus, $o\left(x\right)=1$1 or 5 or 25.

To show every non-identity element of G has order 5

Now, since $x\ne e\Rightarrow o\left(x\right)\ne 1$, where e is an identity element.

Also, is not cyclic $\Rightarrow o\left(x\right)\ne 25$.

Thus, $o\left(x\right)=5$.

This shows that the order of x, which is a non-identity element of Ghas order 5.

Hence, if Gis a group of order 25, then Gis either cyclic or every non-identity element of Ghas order 5.