# Subgroup structure of groups of order 128

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This article gives specific information, namely, subgroup structure, about a family of groups, namely: groups of order 128.
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## Numerical information on counts of subgroups by order

We note the following:

## Abelian subgroups

### Counts of abelian subgroups and abelian normal subgroups

• Congruence condition on number of subgroups of given prime power order tells us that for any fixed order, the number of subgroups is congruent to 1 mod 2 (i.e., it is odd). Since the non-normal subgroups occur in conjugacy classes whose size is a nontrivial power of 2, the number of normal subgroups is congruent to 1 mod 2. In particular, for orders 2 and 4, since every subgroup of that order is abelian anyway, the congruence condition tells us that the number of abelian subgroups is congruent to 1 mod 2, and so is the number of abelian normal subgroups.
• Congruence condition on number of abelian subgroups of prime-cube order and existence of abelian normal subgroups of small prime power order: The existence statement guarantees the existence of an abelian normal subgroup of order $2^3 = 8$ in a group of order $2^7 = 128$, because $7 > 3(3 - 1)/2$. The congruence condition then tells us that the number of such subgroups is odd.
• Congruence condition on number of abelian subgroups of prime-fourth order and existence of abelian normal subgroups of small prime power order: The existence statement guarantees the existence of an abelian normal subgroup of order $2^4 = 16$ in a group of order $2^7= 128$, because $7 > 4(4-1)/2$. The congruence condition then tells us that the number of such subgroups is odd.
• Abelian-to-normal replacement theorem for prime-square index guarantees that if there exists an abelian subgroup of index $2^2$ (or equivalently, order $2^5 = 32$) in a group of order $2^7 = 128$, then there exists an abelian normal subgroup of order $2^5$.
• It also turns out experimentally that the number of abelian subgroups, if nonzero, is odd, so the congruence condition holds -- however, the general theorem is unclear.
• Note that any subgroup of order 64 is normal, so the number of subgroups of order 64 equals the number of normal subgroups of order 64.
• Congruence condition on number of abelian subgroups of prime index guarantees that if there exists an abelian subgroup of prime index, the number of such subgroups is odd. In fact, for a non-abelian group, the only possibilities for the number of abelian subgroups is 0, 1, or 3.