Show that if (G,*) is a group of order 1681 , G is commutative.

Question

Can someone help me with this?

also 1681 is a square of a prime number(41).Can we say the same statement for p^2? where p is any prime?

Answer 1

This is a standard consequence of the fact that a $p$-group has a nontrivial center. Once you know this, then either $G=Z(G)$, and you are done, or $G/Z(G)$ has order $p$, and so it is cyclic. If $G/Z(G)=\langle \bar{g} \rangle$ then every element of $G$ can be written as $g^az$ for $g$ a preimage of $\bar{g}$, $z \in Z(G)$.

Now take any two elements of $G$, $x$ and $y$. Then $x=g^a z$ and $y= g^b w$. So $$xy = g^a z g^b w = g^a g^b zw = g^{a+b} z'$$ and $$yx = g^b w g^a z = g^b g^a wz = g^{b+a} zw = zy$$

so $xy=yx$ and $G$ is abelian.