(Prove)/Disprove that if group A of order 4 is NOT cyclic, then its elements are their own inverses

Question

I am having trouble seeing the relation between cyclic groups/subgroups and inverses; to my knowledge it's cyclic if it can be generated by a single element that spans it across a larger group, but I fail to see how inverses can be cleverly proved using the definition of cyclicism (being cyclic)

so lets suppose...

A is a group of size $4$, denoted $|A| = 4$ ; now prove that A is not cyclic, and then that every element of A is its own inverse

my idea:

A group of order $n$ is cyclic if and only if it has an element of order $n$, now using this definition we could (maybe) draw the conclusion that if our group is order 4 then it's element can't also have order 4 without being the entire group A itself and thus can't be cyclic (?). But then I'm not sure how inverses tie into this unfortunately....

any and all help is appreciated !!

Answer 1

You have the correct intuition. Now, combine the definition of the order of an element and the definition of the inverse, and you're done.

Answer 2

Hint: You have to convince yourself that, the remaining three elements apart from $e$ have order two because can't have order three neither four.

Answer 3

As is well known, the only groups of order $4$ are $\Bbb Z_4$ and $\Bbb Z_2×\Bbb Z_2$. Since we're ruling out the former, we have the latter, that is, the so called Klein four group, where everything but the identity has order $2$.

Answer 4

I think you can prove it with a easy reasoning. Let $G$ be any group and let $g \in G$, we define the order of $g$ as $o(g):=\min\{n \in \mathbb{N} \mid g^n=e\}$. We have the following easy facts.

• $\quad o(g)\leq |G|$.

• $\quad G$ is cyclic if there exists $g\in G$ such that $o(g)=|G|$.

• $\quad o(g)=1$ if and only if $g=e$.

• $\quad $ If $g\not=e$, then $o(g)=2$ if and only if $g^{-1}=g$.

Now, assume that $|G|=4$ and $G$ is non-cyclic. Thus, all elements of $G$ have order at most $3$.

Suppose that there exists $x \in G \setminus\{e\}$ such that $x\not=x^{-1}$. Then $o(x)>2$ and hence $o(x)=3$. Therefore, $G=\{e,x,x^2,y\}$, where $y\not \in \{e,x,x^2\}$. Since $o(x)=3$ we deduce that $x^{-1}=x^2$ and hence $xy\not=1$. If $xy=x$, then $y=e$, which is impossible. Analogously, if $xy=y$, then $x=e$, which is also impossible. Finally, if $xy=x^2$, then $y=x$, which is again a contradiction. Thus, $x=x^-1$ for all $x \in G\setminus\{e\}$ and the result follows.

You can make an easier proof using that the order of the elements must divide $|G|$. However, I prefered to give this elementary proof.