This page describes additional conditions under which a subgroup property implication can be reversed, viz a weaker subgroup property, namely Normal subgroup (?), can be made to imply a stronger subgroup property, namely central subgroup
View other subgroup property implication-reversing conditions
Let be Division ring (?)s (i.e., skew fields), with properly contained in . Suppose is a normal subgroup of . Then is a central subgroup of , viz is contained in the center of . In particular, this implies that is abelian and hence is a field.
Related facts in group theory
- Normal not implies central
- Central implies normal
- Totally disconnected and normal in connected implies central
- Normal of order equal to least prime divisor of group order implies central
Related facts about division rings
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Any element inside commutes with any element outside
For nonzero elements , we denote by the multiplicative commutator and by the element .
We denote by the map . Here, but is allowed to be zero.
Given: and .
To prove: .
Proof: The key idea is to play off the additive and the multiplicative structure against each other, and use the fact that the map is an automorphism of both the additive and the multiplicative structure.
|Step no.||Assertion/construction||Given data/assumptions used||Previous steps used||Explanation||Commentary|
|1||are all in , so the notations , , and make sense.||.||[SHOW MORE]||The choice of and allows us to play on addition.|
|2||and .||is normal in||Step (1)||[SHOW MORE]|
|3||and||Step (1)||[SHOW MORE]||The multiplicative commutator is not convenient because it is not additive/linear in either variable. So, we rewrite it in terms of , which preserves the additive structure.|
|4||Step (1)||[SHOW MORE]||More manipulation.|
|5||, or equivalently,||Steps (3) and (4)||[SHOW MORE]||More manipulation.|
|6||The assumption would lead to a contradiction, hence we must have .||. Also, every nonzero element is invertible because is a skew field.||Steps (2), (5)||[SHOW MORE]||More manipulation.|
|7||Plugging in the result of step (5) gives .||Step (5)||[SHOW MORE]|
The finishing touch
Now, if is a proper subset of , we will show that is contained inside the center. We already know that every element of commutes with every element of . So it suffices to show that any two elements of commute.
Let . Then take any . Then, . Thus, commutes with both and . Hence must commute with the difference, which is .