Complemented normal implies quotient-powering-invariant: Difference between revisions

Latest revision as of 17:55, 16 February 2013

Statement

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 $p^{t h}$ root), $G / H$ is also powered over $p$ .

Facts used

Proof

Proof using given facts

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

Hands-on proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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 $p$ .

Proof:

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation
1	Let $φ : G \to K$ 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 $G / H ≅ K$ .	$H$ 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 $φ (x)^{p} = φ (x^{p}) = φ (g) = g$ . Thus, $φ (x)$ is an element whose $p^{t h}$ power is $g$ . Since $x$ is the unique element of $G$ such that $x^{p} = g$ , we must have $φ (x) = x$ , forcing $x \in K$ .
4	$G / H$ is powered over $p$ .		Steps (1), (3)	Step (1) says that $G / H ≅ K$ . Step (3) says that $K$ is powered over $p$ . Thus, $G / H$ is powered over $p$ .

@@ Line 2: / Line 2: @@
 Suppose <math>G</math> is a [[group]] and <math>H</math> is a [[complemented normal subgroup]] of <math>G</math> (i.e., there exists a permutable complement to <math>H</math> in <math>G</math>, i.e., <math>G</math> is an [[internal semidirect product]] involving <math>H</math>). Then, <math>H</math> is a [[quotient-powering-invariant subgroup]] of <math>G</math>: for any [[prime number]] <math>p</math> such that <math>G</math> is [[group powered over a set of primes|powered over]] <math>p</math> (i.e., every element of <math>G</math> has a unique <math>p^{th}</math> root), <math>G/H</math> is also powered over <math>p</math>.
+==Facts used==
+# [[uses::Complemented normal implies endomorphism kernel]]
+# [[uses::Endomorphism kernel implies quotient-powering-invariant]]
 ==Proof==
+===Proof using given facts===
+The proof follows directly from Facts (1) and (2).
+===Hands-on proof===
 {{tabular proof format}}
@@ Line 20: / Line 31: @@
 | 2 || For <math>g \in K</math>, there exists a unique <math>x \in G</math> such that <math>x^p = g</math>. || || <math>G</math> is powered over <math>p</math>. || ||
 |-
-| 3 || The <math>x</math> obtained in Step (2) is actually an element of <math>K</math>. In other words, for <math>g \in K</math>, there exists a unique <math>x \in K</math> such that <math>x^p = g</math>. So, <math>K</math> is powered over <math>p</math>. || || || Step (2) || We have that <math>\varphi(x)^p = \varphi(x^p) = \varphi(g) = g</math>. Thus, <math>\varphi(x)</math> is an element whose <math>p^{th}</math> power is <math>g</math>. Since <math>x</math> is the unique element of <math>G</math> such that <math>x^p = g</math>, we must have <math>\varphi(x) = x</math>, forcing <math>x \in K</math>.
+| 3 || The <math>x</math> obtained in Step (2) is actually an element of <math>K</math>. In other words, for <math>g \in K</math>, there exists a unique <math>x \in K</math> such that <math>x^p = g</math>. So, <math>K</math> is powered over <math>p</math>. || || || Steps (1), (2) || We have that <math>\varphi(x)^p = \varphi(x^p) = \varphi(g) = g</math>. Thus, <math>\varphi(x)</math> is an element whose <math>p^{th}</math> power is <math>g</math>. Since <math>x</math> is the unique element of <math>G</math> such that <math>x^p = g</math>, we must have <math>\varphi(x) = x</math>, forcing <math>x \in K</math>.
 |-
-| 4 || <math>G/H</math> is powered over <math>p</math>. || || || Steps (1), (3)|| Step (1) says that <math>G/H \cong K</math>. Step (3) says that <math>G/H</math> is powered over <math>p</math>. Thus, <math>K</math> is powered over <math>p</math>.
+| 4 || <math>G/H</math> is powered over <math>p</math>. || || || Steps (1), (3)|| Step (1) says that <math>G/H \cong K</math>. Step (3) says that <math>K</math> is powered over <math>p</math>. Thus, <math>G/H</math> is powered over <math>p</math>.
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