**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.1 Exercise 2.1.12**

Let $A$ be an abelian group and fix $n \in \mathbb{Z}^+$. Prove that the following subsets are subgroups of $A$.

(1) $M = \{ a^n \ |\ a \in A \}$,

(2) $M = \{ a \in A \ |\ a^n = 1 \}$.

Solution:

(1) Note that $M$ is not empty since $1 = 1^n$. Now suppose $x,y \in M$. Then we have $x = g^n$, $y = h^n$ for some $g,h \in A$. Then, using Exercise 1.1.24, we have $$xy^{-1} = g^nh^{-n} = (gh^{-1})^n;$$ thus $xy^{-1} \in M$. By the subgroup criterion, $M \leq A$.

(2) Note that $M$ is not empty since $1 = 1^n$. Now suppose $x,y \in M$. Then $x^n = y^n = 1$. We have $$(xy^{-1})^n = x^n(y^n)^{-1} = 1,$$ so that $xy^{-1} \in M$. By the subgroup criterion, $M \leq A$.