Subgroups of p p -groups

Algebra Level 2

Let G G be a group of order p 5 , p^5, where p p is prime. Consider the following statements:

I. G G has a normal subgroup of order p . p.
II. G G has a normal subgroup of order p 2 . p^2.
III. G G has a normal subgroup of order p 3 . p^3.
IV. G G has a normal subgroup of order p 4 . p^4.

How many of the statements I-IV is/are always true for any G G of order p 5 p^5 ?

Terminology : The order of a finite group is the number of elements in the group.

1 3 2 4 0

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1 solution

Patrick Corn
Aug 1, 2016

It is a fact that a group of order p n p^n has normal subgroups of order p k p^k for any k n . k \le n. This can be proved by induction.

The base case n = 1 n=1 is clear. Now suppose the result has been established for all group of order < p n , <p^n, and let G G be a group of order p n . p^n. There are two key facts. The first is that the center Z ( G ) Z(G) is nontrivial (see the p-groups wiki for details). Then the quotient group G / Z ( G ) G/Z(G) is a smaller p p -group, so has normal subgroups of every possible size by the inductive hypothesis. The second key fact is that the normal subgroups of G / Z ( G ) G/Z(G) are in one-to-one correspondence with normal subgroups of G G containing Z ( G ) , Z(G), by the third isomorphism theorem . So if Z ( G ) = p k , |Z(G)|=p^k, this shows that there are normal subgroups of order p k , p k + 1 , , p n p^k,p^{k+1},\ldots,p^n containing Z ( G ) . Z(G).

It remains to be shown that the center contains normal subgroups of every possible size. Note that every subgroup of the center is normal in G , G, so it is only necessary to find subgroups of every possible size. And so the result follows from the structure theorem for abelian groups , and induction: any abelian group of order p n p^n is isomorphic to a product of groups of the form Z p i {\mathbb Z}_{p^i} , and a subgroup of order p n 1 p^{n-1} can be obtained simply by changing one of the Z p i {\mathbb Z}_{p^i} to Z p i 1 . {\mathbb Z}_{p^{i-1}}. That subgroup will have subgroups of every possible order by the inductive hypothesis.

(Note that the second paragraph of the proof is necessary: if G G is abelian, the center equals G , G, so you can't use the original inductive hypothesis to get subgroups of it; you need to know the theorem is true for abelian groups.)

So the statements in the problem are all always true, and the answer is 4 . \fbox{4}.

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