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Complemented normal implies quotient-powering-invariant



Suppose G is a group and H is a complemented normal subgroup of G (i.e., there exists a permutable complement to H in G, i.e., G is an internal semidirect product involving H). Then, H is a quotient-powering-invariant subgroup of G: for any prime number p such that G is powered over p (i.e., every element of G has a unique p^{th} root), G/H is also powered over p.

Facts used


Proof using given facts

The proof follows directly from Facts (1) and (2).

Hands-on proof

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Given: A group G, a normal subgroup H of G with permutable complement K. G is powered over a prime number p.

To prove: G/H is powered over p.


Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let \varphi:G \to K be the retraction that sends an element of G to the unique element of K in its coset with respect to H. This corresponds to the quotient map G \to G/H and G/H \cong K. H is normal in G and is complemented, with complement K.
2 For g \in K, there exists a unique x \in G such that x^p = g. G is powered over p.
3 The x obtained in Step (2) is actually an element of K. In other words, for g \in K, there exists a unique x \in K such that x^p = g. So, K is powered over p. Steps (1), (2) We have that \varphi(x)^p = \varphi(x^p) = \varphi(g) = g. Thus, \varphi(x) is an element whose p^{th} power is g. Since x is the unique element of G such that x^p = g, we must have \varphi(x) = x, forcing x \in K.
4 G/H is powered over p. Steps (1), (3) Step (1) says that G/H \cong K. Step (3) says that K is powered over p. Thus, G/H is powered over p.