Group Properties and p-power Order

The question describes a group \(G\) where every element \(g\) has an order that is a power of a prime number \(p\). This property is characteristic of \(p\)-groups (if finite) or groups where every element has \(p\)-power order (if potentially infinite).

Subgroup Statement Analysis

Statement 1 says: If \(|G| = p^6\), then \(G\) has a subgroup of index \(p^2\).

If \(|G| = p^6\), \(G\) is a finite group of order \(p^6\), which is a finite \(p\)-group.

A fundamental theorem about finite \(p\)-groups states that a group of order \(p^k\) has subgroups of every order \(p^i\) for \(0 \le i \le k\). Having a subgroup of order \(p^i\) is equivalent to having a subgroup of index \(p^{k-i}\).

For \(|G| = p^6\), \(k=6\). We are looking for a subgroup \(H\) with index \(|G:H| = p^2\). The order of such a subgroup would be \(|H| = \frac{|G|}{|G:H|} = \frac{p^6}{p^2} = p^4\).

Since \(G\) is a finite \(p\)-group of order \(p^6\), it must have a subgroup of order \(p^4\). This subgroup has index \(p^2\).

Thus, Statement 1 is TRUE.

Normal Subgroup Statement Analysis

Statement 2 says: If \(|G| = p^6\), then \(G\) has at least five normal subgroups.

Again, \(|G| = p^6\) implies \(G\) is a finite \(p\)-group.

A property of finite \(p\)-groups of order \(p^k\) (\(k \ge 1\)) is that they have normal subgroups of every order \(p^i\) for \(0 \le i \le k\).

For \(|G| = p^6\), there exist normal subgroups of orders \(p^0, p^1, p^2, p^3, p^4, p^5, p^6\). These correspond to normal subgroups with orders:

• Order \(p^0 = 1\): The trivial subgroup \(\{1\}\). This is always a normal subgroup.
• Order \(p^1\).
• Order \(p^2\).
• Order \(p^3\).
• Order \(p^4\).
• Order \(p^5\).
• Order \(p^6\): The group \(G\) itself. This is always a normal subgroup.

These are 7 distinct possible orders for normal subgroups. Since there is at least one normal subgroup for each order, a group of order \(p^6\) has at least 7 normal subgroups.

Having at least 7 normal subgroups certainly means having at least five normal subgroups.

Thus, Statement 2 is TRUE.

Center Statement Analysis

Statement 3 says: Center of \(G\) can be infinite.

The condition is that every element in \(G\) has \(p\)-power order. The group \(G\) is not necessarily finite.

Consider the Prüfer group \(C_{p^\infty}\). This is an infinite group where every element has order \(p^k\) for some non-negative integer \(k\).

The Prüfer group \(C_{p^\infty}\) is an abelian group. In any abelian group, the center \(Z(G)\) is equal to the group \(G\) itself.

For \(G = C_{p^\infty}\), the center \(Z(G) = C_{p^\infty}\), which is an infinite group.

Therefore, there exists a group \(G\) satisfying the given condition (every element has \(p\)-power order) whose center is infinite.

Thus, Statement 3 is TRUE.

Exactly Six Normal Subgroups Analysis

Statement 4 says: There exists \(G\) with \(|G| = p^6\) such that \(G\) has exactly six normal subgroups.

From our analysis of Statement 2, we established that any finite \(p\)-group \(G\) of order \(p^6\) must have normal subgroups of orders \(p^0, p^1, p^2, p^3, p^4, p^5, p^6\). This guarantees the existence of at least one normal subgroup for each of these 7 distinct orders.

This implies that any group of order \(p^6\) must have at least 7 normal subgroups (the trivial group, the group itself, and at least one for each intermediate order \(p^i\), \(1 \le i \le 5\)).

It is impossible for a group of order \(p^6\) to have exactly six normal subgroups because it is guaranteed to have at least seven.

Thus, Statement 4 is FALSE.

False Statement Conclusion

We have analyzed each statement:

• Statement 1: If \(|G| = p^6\), then \(G\) has a subgroup of index \(p^2\). (TRUE)
• Statement 2: If \(|G| = p^6\), then \(G\) has at least five normal subgroups. (TRUE)
• Statement 3: Center of \(G\) can be infinite. (TRUE)
• Statement 4: There exists \(G\) with \(|G| = p^6\) such that \(G\) has exactly six normal subgroups. (FALSE)

The question asks which of the statements is FALSE.

The statement that is FALSE is: There exists \(G\) with \(|G| = p^6\) such that \(G\) has exactly six normal subgroups.