Central implies normal satisfying the subgroup-to-quotient powering-invariance implication
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., central subgroup) must also satisfy the second subgroup property (i.e., normal subgroup satisfying the subgroup-to-quotient powering-invariance implication)
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Suppose is a group and is a central subgroup of . Note that central implies normal, so is normal. Then, is a normal subgroup satisfying the subgroup-to-quotient powering-invariance implication in : if is a prime number such that both and are -powered, so is the quotient group .
- is a central subgroup of , i.e., every element of commutes with every element of .
- is a powering-invariant subgroup of , i.e., for any prime such that is powered over , so is .
Proof for original formulation
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Given: A group . A central subgroup of A prime number such that is powered over , i.e., every element of has a unique root in . is also powered over . Let be the quotient map. An element .
To prove: There is a unique element satisfying .
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||Let be such that .||is surjective by definition.|
|2||There exists such that . Remember this .||is powered over (we only use the divisibility aspect here, not the uniqueness).||direct.|
|3||For any , there exists such that .||is powered over .||given-direct.|
|4||For any element of of the form , with as in Step (2) and an arbitrary element of as in Step (3), we have where is the element from Step (2) and is found as in Step (3).||is central.||Steps (2), (3)||Since is central and , we get that .|
|5||is the only element of whose power is .||is powered over (this is where we use the uniqueness).||Step (4)||Note first that since , we get , so . Suppose there is an element such that . Let . Then, , hence is of the form . By Step (4), there exists such that . By the uniqueness part of -powering, we obtain that . Hence, . Thus, , yielding . In other words, any element of whose power is must equal .|
Proof for corollary formulation
This is immediate from the original formulation, and the observation that:
(actually, the converse implication also holds, but that is not necessary for us here).