Complemented normal implies quotient-powering-invariant
Suppose is a group and is a complemented normal subgroup of (i.e., there exists a permutable complement to in , i.e., is an internal semidirect product involving ). Then, is a quotient-powering-invariant subgroup of : for any prime number such that is powered over (i.e., every element of has a unique root), is also powered over .
Proof using given facts
The proof follows directly from Facts (1) and (2).
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Given: A group , a normal subgroup of with permutable complement . is powered over a prime number .
To prove: is powered over .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||Let be the retraction that sends an element of to the unique element of in its coset with respect to . This corresponds to the quotient map and .||is normal in and is complemented, with complement .|
|2||For , there exists a unique such that .||is powered over .|
|3||The obtained in Step (2) is actually an element of . In other words, for , there exists a unique such that . So, is powered over .||Steps (1), (2)||We have that . Thus, is an element whose power is . Since is the unique element of such that , we must have , forcing .|
|4||is powered over .||Steps (1), (3)||Step (1) says that . Step (3) says that is powered over . Thus, is powered over .|