Can you please check my logic and maybe suggest some other strategies for the following problem:
Prove that every group of order $4$ is abelian as follows: Let $G$ be any group of order $ 4$, i.e., $|G| = 4$.
(1) Suppose there exists $a \in G$ such that $o(a) = 4$. Prove that $G$ is abelian.
(2) Suppose that no element of $G$ has order 4. Prove that $\forall x\in G$, $x^2 = 1$.
(3) Suppose that no element of $G$ has order 4. Prove
that $G$ is abelian.
What I got so far:
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(1) If there exists $a \in G$ such that $o(a) = 4$,
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Case 1: $a\cdot a=b$. Then $a\cdot a\cdot a=c$ and $a\cdot a\cdot a\cdot a=1$. Algebra… G is abelian.
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Case 2: $a\cdot a=c$… $G$ is abelian.
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(2) Let $x\in G$. If $o(x) \neq 4$, we can clarify that an element cannot have an order greater than $4$ in a group of order $4$ and that the only element that has an order of $1$ is $1$. Therefore the other three elements must have an order of $2$, so $x^2=2$ for all $x \in G$.
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(3) No ideas yet 🙁
Best Answer
Another strategy to 1)
If there exist some $a \in G$ such that $O(a) = 4$ then $G = \{e,a,a^{2},a^{3} \}$
Thus $c,d \in G \implies c= a^{i}, d = a^{j} \implies c\cdot d = a^{i} \cdot a^{j} = a^{i+j} = a^{j} \cdot a^{i} = d\cdot c$
3) Suppose there no exist $a \in G$ such that $O(a) = 4$
Then if $a \in G, a \not= e \implies O(a) = 2$
Let $a,b \in G$ then $a \cdot b \in G \implies (a\cdot b)^{2} = e \implies (a\cdot b)(a\cdot b) = e \implies a\cdot b \cdot a \cdot b = e $
Thus $a \cdot (a\cdot b \cdot a \cdot b) \cdot b = a \cdot e \cdot b = a \cdot b \implies a^{2}\cdot b \cdot a \cdot b^{2} = a\cdot b \implies e \cdot b \cdot a \cdot e = a \cdot b$
$ \implies b \cdot a = a \cdot b $
Then $G$ is an abelian group.