Congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group
From Groupprops
Contents
Statement
Suppose is a group of prime power order. Suppose is a collection of subgroups of . For any subgroup of , denote by the number of subgroups of that are in . Then, if , we have:
If , we have:
.
Related facts
Similar facts
- Congruence condition relating number of normal subgroups containing minimal normal subgroups and number of normal subgroups in the whole group
- Congruence condition on number of subgroups of given prime power order
Facts used
- Equivalence of definitions of maximal subgroup of group of prime power order (specifically, nilpotent implies every maximal subgroup is normal, and prime power order implies nilpotent, so all maximal subgroups are normal in groups of prime power order).
- Fourth isomorphism theorem
- Formula for number of maximal subgroups of group of prime power order
Proof
Note that the case follows directly from the case, so we restrict attention to the case.
Given: is a group of prime power order. is a collection of subgroups of . For any subgroup of , denote by the number of subgroups of that are in .
To prove: , where the summation is over maximal subgroups of . Note that .
Proof:
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | Let be the set of ordered pairs where and is a maximal subgroup of containing . | ||||
2 | . | Step (1) | [SHOW MORE] | ||
3 | is the sum, over , of the number of maximal subgroups of containing . | Step (1) | Step-direct. | ||
4 | For any subgroup of , the maximal subgroups of containing are precisely the same as the maximal subgroups of containing the normal closure of in . | Fact (1) | [SHOW MORE] | ||
5 | For any subgroup of , the maximal subgroups of containing correspond to maximal subgroups of where is the normal closure of in . | Fact (2) | Step (4) | Step-fact combination direct. | |
6 | For any proper subgroup of , the number of maximal subgroups of containing is congruent to 1 mod . | Fact (3) | Steps (4), (5) | [SHOW MORE] | |
7 | . Note that . | Steps (3), (6) | [SHOW MORE] | ||
8 | Steps (2), (7) | Step-combination direct. |
References
Journal references
- A contribution to the theory of groups of prime power order by Philip Hall, Proceedings of the London Mathematical Society, ISSN 1460244X (online), ISSN 00246115 (print), Page 29 - 95(Year 1934): In this paper, Philip Hall summarizes and extends many of the known results about groups of prime power order.^{}^{More info}
- Counting abelian subgroups of p-groups: a projective approach by Marc Konvisser and David Jonah, Journal of Algebra, ISSN 00218693, Volume 34, Page 309 - 330(Year 1975): ^{PDF (ScienceDirect)}^{More info}