Every normal subgroup satisfies the quotient-to-subgroup 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., quotient-powering-invariant subgroup) must also satisfy the second subgroup property (i.e., powering-invariant subgroup)
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Suppose is a group and is a normal subgroup of . Suppose is a prime number such that is powered over (every element of has a unique root). Suppose that the quotient group is also powered over . Then, is also powered over .
The idea is to apply the uniqueness of roots in the quotient group to the case of the roots of the identity element of the quotient group.
Proof details for original formulation
Given: A group , a normal subgroup of . A prime such that both and are powered over . An element .
To prove: There is a unique element such that .
Proof: Let be the quotient map.
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||There exists a unique element such that .||is powered over , .||Direct.|
|2||The of Step (1) is in .||is powered over , .||Step (1)||Since is powered over , there is a unique element whose power is the identity element of . This unique element must therefore be the identity element of . But we also have that is the identity element of since . This forces that is the identity element of , so .|
Proof for corollary formulation
This is direct from the original formulation once we note that:
is powered over and is quotient-powering-invariant in is powered over .