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If $G$ is a finite $p$-group of order $p^n$, then it is well known that for ($1\leq m\leq n$), number of subgroups of order $p^m$ is $1$(mod $ p$).

Question: Is it true that number of subgroups of order $p^m$, which are isomorphic within themselves, is $0$(mod $ p$) or $1$(mod $p$).

It looks to be true for groups of order $p^2$, $p^3$.

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What does "isomorphic within themselves" mean? – Igor Rivin Sep 9 '11 at 11:20
Given a subgroup $L$ of $G$, consider the set $X=\{H\leq G \colon H\cong L\}$. Does the cardinality of this set is $0$ (mod $p$) or $1$ (mod $p$)? For groups of order $p^2$, and $p^3$, it looks to be true. – joseph Sep 9 '11 at 13:45
up vote 4 down vote accepted

This is just a matter of searching for a counterexample!

You will find that in GAP or Magma, SmallGroup(81,7) has four subgroups of order 27, two of which are isomorphic, but the other two are not isomorphic to any others.

It is possible that the answer might be yes for abelian groups.

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Wrong for groups of order $3^4$ or $5^4$.

Here is Magma code (can be tested in ):

p := 5;

a := 5;

ord := p^a;

n := NumberOfSmallGroups(ord);

for i in [1..n] do

G := SmallGroup(ord,i);

ST := [SubgroupLattice(G)[i] : i in [1..#SubgroupLattice(G)]];

print [&+[Integers()!(Order(G)/Order(Normaliser(G,H))) : H in ST | IsIsomorphic(H,K)] mod p : K in ST];

end for;

Testing groups of order $5^5$, we see that the number in question is also not necessarily congruent to -1, 0 or +1 modulo p.

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