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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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An extraspecial p-group has order p1+2n for some n1

Statement

Let P be an extraspecial p-group. Then P=p1+2n for some integer n1, and P/Z(P)=p2n. In particular no extraspecial group has order p2m, and none has order p.

Facts & Assumptions

Given: An extraspecial p-group P.

[F1]

Z(G):={zG:zg=gz for every gG} (The center Z(G) of a group).

[L1]

Every extraspecial p-group is an internal central product of n1 nonabelian subgroups of order p3 with pairwise intersections Z(P), and P=p1+2n (Every extraspecial p-group is an internal central product of nonabelian subgroups of order p3).

[L2]

For a finite p-group P the following are equivalent: P is extraspecial; P is nonabelian, Z(P)=p and P/Z(P) is elementary abelian; P is nonabelian and Z(P)=P=Φ(P) has order p (Three equivalent descriptions of an extraspecial p-group).

[L3]

For a finite group G and HG, G=[G:H]H (Lagrange's theorem: G=[G:H]H for every subgroup H of a finite group G).

[L4]

[G:H]:=G/H, the number of left cosets of H in G (The coset set G/H and the index [G:H] of a subgroup).

Proof

technique · direct
1.1

The decomposition theorem writes P as an internal central product of n1 nonabelian subgroups of order p3 and gives P=p1+2n.

L1
1.2

The centre of P has order p.

F1L2
2.1

Lagrange applied to Z(P)P gives P/Z(P)=P/p=p2n.

L3L4step 1.1step 1.2
3.1

Since 1+2n is odd, no extraspecial group has order an even power of p; and n1 excludes the order p, which corresponds to n=0.

step 1.1step 2.1

Remarks

The exponent n is determined by the order and therefore by the group, so it can be used as an invariant even though the decomposition producing it is not unique. That n1 is what the nonabelian clause of the definition buys: an abelian group with a centre of order p would be the cyclic group of order p, of order p1+20.

Depends on

Used by

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Sources