Assertion: abelian group's subgroup | /r/HasProperty | proper
Facets:
value: in this case
type: TEMPORAL
Source sentences:
text: The subgroup of false witnesses is, in this case, not proper; it is the entire group of multiplicative units modulo 561, which consists of 320 residues.
text: The alternating group of degree n is always a normal subgroup, a proper one for n ≥ 2 and nontrivial for n ≥ 3; for n ≥ 3 it is in fact the only non-identity proper normal subgroup of S n , except when n = 4 where there is one additional such normal subgroup, which is isomorphic to the Klein four group.
text: A proper subgroup of a group G is a subgroup H which is a proper subset of G (that is, H ≠ G).
text: Note that b is of order p, so the subgroup of G generated by b contains p elements and thus is a proper subset of C G (b), because C G (b) includes all elements of this subgroup and the center which does not contain b but at least p elements.