Divisibility is quotient-closed

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Statement

Suppose G is a group and H is a normal subgroup of G, so that G/H is the quotient group. Then, if n is a natural number such that G is n-divisible (i.e., every element of G has a n^{th} root in G, then so is G/H, i.e., every element of G/H has a n^{th} root in H.

Proof

Given: A group G and a natural number n such that every element of G has a n^{th} root in G, a normal subgroup H with quotient group G/H. An element a \in G/H.

To prove: There exists b \in G/H such that b^n = a.

Proof: Let \varphi:G \to G/H be the quotient map and let g \in G be such that \varphi(g) = a. Since G is n-divisible, there exists h \in G such that h^n = g. Let b = \varphi(h). Then, b^n = (\varphi(h))^n = \varphi(h^n) = \varphi(g) = a, so b is as desired.