Every normal subgroup satisfies the quotient-to-subgroup powering-invariance implication

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., quotient-powering-invariant subgroup) must also satisfy the second subgroup property (i.e., powering-invariant subgroup)
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Statement

Original formulation

Suppose G is a group and H is a normal subgroup of G. Suppose p is a prime number such that G is powered over p (every element of G has a unique p^{th} root). Suppose that the quotient group G/H is also powered over p. Then, H is also powered over p.

Corollary formulation

Suppose G is a group and H is a normal subgroup of G that is a quotient-powering-invariant subgroup: if G is powered over a prime p (i.e., every element of G has a unique p^{th} root), so is the quotient group G/H.

Then, H is also a powering-invariant subgroup of H: if G is powered over a prime p, so is H.

Proof

Proof idea

The idea is to apply the uniqueness of p^{th} roots in the quotient group to the case of the p^{th} roots of the identity element of the quotient group.

Proof details for original formulation

Given: A group G, a normal subgroup H of G. A prime p such that both G and G/H are powered over p. An element g \in H.

To prove: There is a unique element x \in H such that x^p = g.

Proof: Let \pi:G \to G/H be the quotient map.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 There exists a unique element x \in G such that x^p = g. G is powered over p, x \in H \le G. Direct.
2 The x of Step (1) is in H. G/H is powered over p, g \in H. Step (1) Since G/H is powered over p, there is a unique element whose p^{th} power is the identity element of G/H. This unique element must therefore be the identity element of G/H. But we also have that \pi(x)^p = \pi(x^p) = \pi(g) is the identity element of G/H since g \in H. This forces that \pi(x) is the identity element of G/H, so x \in H.

Proof for corollary formulation

This is direct from the original formulation once we note that:

G is powered over p and H is quotient-powering-invariant in G \implies G/H is powered over p.