Understanding Groups of Order pⁿ

The question asks about the properties of a group G with order $|G| = p^n$, where p is a prime number and n is an integer greater than 1. Groups whose order is a power of a prime are called p-groups. These groups have special properties that distinguish them from other types of groups.

Let's analyze each given option based on known results in group theory regarding p-groups.

Centre of a Group of Order pⁿ

The first option states that the centre of G has at least two elements. The centre of a group G, denoted by Z(G), is the set of elements that commute with every other element in G. For any non-trivial p-group, a fundamental theorem states that its centre is non-trivial. This means that if $|G| = p^n$ and n > 0, then $|Z(G)| > 1$.

In this case, the order of the group is $|G| = p^n$ with n > 1. Since n > 1, the order is $p^n \ge p^2$. As p is a prime, $p \ge 2$. Thus, $|G| \ge 2^2 = 4$, which means G is a non-trivial group.

For any non-trivial p-group G, the order of its centre, $|Z(G)|$, must be a power of p, say $p^k$, where $k \ge 1$. This implies $|Z(G)| \ge p$. Since p is a prime number, $p \ge 2$. Therefore, $|Z(G)| \ge 2$. This confirms that the centre of G has at least two elements.

Thus, the first statement is true.

Group is Always Abelian?

The second option suggests that a group G of order $p^n$ (n > 1) is always an Abelian group. This is not true.

• Groups of order $p^2$ are always Abelian. However, for n > 2, there exist non-Abelian p-groups.
• A common example is the quaternion group Q₈, which has order 8 = $2^3$ (here p=2, n=3). The quaternion group is not Abelian because, for instance, $i \cdot j = k$ but $j \cdot i = -k$.

Therefore, the second statement is false.

Normal Subgroups of a Group of Order pⁿ

The third option claims that G has exactly two normal subgroups, meaning G is a simple group. A simple group is a group whose only normal subgroups are the trivial subgroup ({e}) and the group itself (G).

As discussed earlier, for a non-trivial p-group G, the centre Z(G) is non-trivial. The centre Z(G) is always a normal subgroup of G. Since n > 1, G is non-trivial, so Z(G) is also non-trivial. If Z(G) is not equal to G (i.e., G is non-Abelian), then Z(G) is a normal subgroup distinct from {e} and G. If G is Abelian, then every subgroup is normal. For n > 1, G has subgroups other than {e} and G (e.g., a subgroup of order p). These subgroups are normal in an Abelian group.

In fact, it can be shown that any non-trivial p-group has a normal subgroup of order $p^k$ for every $1 \le k \le n$. This means a group of order $p^n$ with n > 1 must have normal subgroups other than just {e} and G. For instance, it has a normal subgroup of order p (contained within the center or a larger normal subgroup) and normal subgroups of other orders up to $p^{n-1}$.

Thus, a group of order $p^n$ with n > 1 cannot be simple. The third statement is false.

Isomorphism with Other Groups of the Same Order

The fourth option suggests that if H is any other group of order $p^n$, then G is isomorphic to H. This is generally not true.

• Groups of the same order are not necessarily isomorphic. While there is only one group (up to isomorphism) of order p (the cyclic group C_p), for order $p^2$ there are two non-isomorphic groups (C_p² and C_p × C_p).
• For higher powers of p (n > 2), the number of non-isomorphic groups of order $p^n$ increases rapidly. For example, there are 5 non-isomorphic groups of order 8 ($2^3$).

Therefore, the fourth statement is false.

Based on the analysis, only the first statement is true.

The final answer is $\boxed{\text{Centre of G has at least two elements}}$.