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  1. In a group $(G, \circ)$, $(a\circ b)^3=a^3 \circ b^3$ and $(a \circ b)^5=a^5 \circ b^5$ $ \forall$ $ a, b \in G$, prove that the group is abelian.

My Solution: $(a^3 \circ b^3) \circ (a\circ b)^2 =a^5 \circ b^5 \implies a^3 \circ b^3 \circ (a \circ b)^2 = a^2 \circ (a \circ b)^3 \circ b^2 \implies a \circ b^3 \circ (a \circ b)^2 = (a \circ b)^3 \circ b^2 \implies (a\circ b) \circ b^2 \circ (a \circ b)^2 = (a \circ b)^3 \circ b^2 \implies b^2 \circ (a \circ b)^2 = (a \circ b)^2 \circ b^2$

As $a \circ b= c \in G \forall a, b \in G$, therefore: $b^2 \circ c^2 = c^2 \circ b^2 \implies c \circ b^2 \circ c^2 \circ b= c^3 \circ b^3 \implies (c\circ b)\circ ( b\circ c) \circ (c\circ b) = (c\circ b)^3 \implies b\circ c = c\circ b$, i.e the Group is abelian.

Is this process at all correct? Please verify.

  1. If each element of a group, [say, $(G, *)$ ]except the identity be of order 2, prove that it is abelian.

My Solution:

$o(a)=2 \implies a^2=e$ $ \forall$ $ a \in G-\{e\}$ Hence, $ a^2=b^2=e$ $[ b\in G -\{e\}] $$\implies a^2*b^2=e=(a*b)^2 $[as $a*b \in G$] $ \implies a*(a*b)*b=(a*b)*(a*b)\implies a*(a*b)*b=a*(b*a)*b \implies a*b=b*a$ ,i.e the Group is abelian.

Kindly go through the above two solutions and make corrections wherever necessary. I am learning the basics by myself, so a little help goes a long way.

Subhasis Biswas
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1 Answers1

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The second solution is good. I feel a little bit uncomfortable for the first one. To prove a group is abelian, I often write let $a,b\in G$ at the beginning, and then get $ab=ba$ at the end. Using your argument, I would write like this:

Let $b,c\in G$. Let $a=cb^{-1}$. Then by your first line, we have $b^{2}(ab)^{2}=(ab)^{2}b^{2}$. Since $c=ab$, we have $b^{2}c^{2}=c^{2}b^{2}$. Then by your second line argument, we have $bc=cb$. So $G$ is abelian.

I would also put let $a,b\in G$ at the beginning of the second proof, so that people know what $a$ and $b$ are.

Delong
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