[Math] 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.

abstract-algebra

If $G$ is a group of order $25$

Prove that either $G$ is cyclic or else evry nonIdentyt element of $G$ has order 5.

Here are few theorems that I am trying to work with might be irrelevant thougth

Thm1
Let S be a nonempty subset of group $G$ . Let $\langle S \rangle$ be set of all possible products, in every order, of elements of $S$ and their inverses

$\Rightarrow$

  1. $\langle S \rangle$ is a subgroup of set $S$
  2. $H$ is a subgroup of $G$ when $S \subset H \Rightarrow \langle S \rangle \leq H$

Thm2 Every subgroup of a cyclic group is itself cyclic

Thm3 Let $G$ be a group and let $a\in G$

  1. If $a$ has infinite order then $\langle a \rangle$ is an infinite subgroup consisting of distinct $k a$, $k \in Z$

    1. If $a$ has order $n$ then $\langle a \rangle$ is a subgroup of order $n $
      and $\langle a \rangle=\{ 0,1a,2a, \dots , \dots , (n-1)a \}$

I don't know where to go from there or if it can be done from scratch easily.

Best Answer

Hint: By Lagrange's theorem, the order of any element divides $25$. There is exactly one element of order $1$. Thus, we have two cases: either the group contains an element of order $25$, or it contains no such element.

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