Groups of Order p²: Moving Beyond Prime Order

 After teaching groups of prime order, students naturally ask: "What about p²?" This is where things get interesting! While prime order groups are always cyclic, p² order groups have exactly TWO possibilities.

Theorem: Let G be a group of order p² where p is prime. Then G is abelian and isomorphic to either:

Z_{p²} (cyclic group), or

Z_p × Z_p (direct product)

Proof : We use the class equation and properties of p-groups.

Key Fact: Every p-group has a non-trivial center.

For |G| = p², the center Z(G) has order dividing p².

Since Z(G) is non-trivial, |Z(G)| ∈ {p, p²}.

Case 1: If |Z(G)| = p², then Z(G) = G, so G is abelian.

Case 2: If |Z(G)| = p, then |G/Z(G)| = p²/p = p (prime).

By our previous theorem, G/Z(G) is cyclic.

But if G/Z(G) is cyclic, then G must be abelian!

Conclusion: In both cases, G is abelian. 


The Two Types

Type 1: Cyclic Group Z_{p²}

Example: Z₉ = {0, 1, 2, 3, 4, 5, 6, 7, 8}

Properties:

Has elements of order p² (the generators)

Number of generators = φ(p²) = p(p-1)

Exactly one subgroup of each order dividing p²

For Z₉:

Elements of order 9: {1, 2, 4, 5, 7, 8} (6 elements = φ(9))

Elements of order 3: {3, 6}

Identity has order 1


Type 2: Direct Product Z_p × Z_p

Example: Z₃ × Z₃ = {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2)}

Properties:

Every non-identity element has order p (no elements of order p²!)

Has more than one subgroup of order p

Can be viewed as a vector space over Z_p


For Z₃ × Z₃:

All non-identity elements have order 3

Has 4 subgroups of order 3:

⟨(1,0)⟩ = {(0,0), (1,0), (2,0)}

⟨(0,1)⟩ = {(0,0), (0,1), (0,2)}

⟨(1,1)⟩ = {(0,0), (1,1), (2,2)}

⟨(1,2)⟩ = {(0,0), (1,2), (2,1)}


How to Distinguish Them

Quick Test: Find the maximum order of elements.

If max order = p²: Group is Z_{p²} (cyclic)

If max order = p: Group is Z_p × Z_p


Example: Determine the structure of a group G with |G| = 25.

If G has an element of order 25, then G ≅ Z₂₅.

If all non-identity elements have order 5, then G ≅ Z₅ × Z₅.


Subgroup Structure

Subgroups of Z_{p²}

By Lagrange, possible orders: 1, p, p²


One subgroup of order 1: {e}

Exactly one subgroup of order p

One subgroup of order p²: G itself


Subgroups of Z_p × Z_p

Possible orders: 1, p, p²


One subgroup of order 1: {e}

p + 1 subgroups of order p

One subgroup of order p²: G itself


Example for Z₃ × Z₃:

Has 3 + 1 = 4 subgroups of order 3 (listed above).


The Fundamental Theorem Connection

This result is a special case of the Fundamental Theorem of Finite Abelian Groups:

Every finite abelian group is a direct product of cyclic groups of prime power order.

For order p², the only partitions are:


p² = p² → gives Z_{p²}

p² = p · p → gives Z_p × Z_p


Practice Problems

Question: How many non-isomorphic groups of order 49 exist?

A) 1 B) 2 C) 3 D) 7

Answer: B) 2


Question: A group G of order 16 has all non-identity elements of order 2. Can G be cyclic?

A) Yes

B) No

C) Only if abelian

Answer: B) No


Concrete Examples

Groups of Order 4

4 = 2², so we have exactly two groups:


Z₄: Cyclic, e.g., {0, 1, 2, 3} under addition mod 4

Z₂ × Z₂: Klein four-group V₄ = {e, a, b, ab}


Groups of Order 9

9 = 3², so we have exactly two groups:


Z₉: Cyclic

Z₃ × Z₃: Direct product


Groups of Order 25

25 = 5², so we have exactly two groups:


Z₂₅: Cyclic

Z₅ × Z₅: Direct product

Understanding these two types deeply prepares you for more complex group classifications. On my YouTube channel "Maths mastery with Dr. Upasana P Taneja," I work through many examples showing how to identify and work with these structures!


Two types, infinite applications – that's the power of p²!

Dr. Upasana Pahuja Taneja

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