Group of order $p^2$

Question

Let $G$ be a group of order $p^2$ where $p$ is a prime. I want to show that either $G$ is cyclic or is the product of two cyclic groups of order $p$.

After some work, the problem reduces to showing the following:

Let $x,y \in G$ be two elements of order $p$ and $y \notin \langle x\rangle$.

Need to show that

1) $\langle x\rangle$ is normal

2) $\langle x\rangle \cap \langle y\rangle={e}$

3) $\langle x\rangle \langle y\rangle = G$

Anyone can give some hints? Thanks!

Answer 1

Assuming (1):

2. Let $g \in \langle x\rangle \cap \langle y\rangle$. Then $g=y^k$ for some $k$ with $0\le k < p$. If $k>0$, then $\langle g\rangle = \langle y\rangle$ and $y \in \langle g\rangle \subseteq \langle x\rangle \cap \langle y\rangle \subseteq \langle x\rangle$, contrary to the hypothesis. Therefore, $k=0$ and $g=e$.

3. $\langle x\rangle \langle y\rangle$ contains all elements of the form $x^i y^j$ with $0 \le i,j < p$. These elements are all different because of (2). Therefore, $\langle x\rangle \langle y\rangle$ has at least $p^2$ elements and so must be equal to $G$.

Answer 2

$Z(G)$ denotes the center of $G$.

1. Prove that if $G/Z(G)$ is cyclic, then $G$ is abelian.
2. Prove that a group of order $p^k$ for prime $p$ has nontrivial center.
3. Use the classification of abelian group to derive the conclusion about the structure of groups of order $p^2$.

Answer 3

Note of caution: I am working in modulo p without explicitly specifying it in some arguments (exponents of elements).

Now, we have two possibilities. If $ G $ has an element of order $ p^2 $, then we are done as this implies that $ G $ is cyclic. Otherwise, $ G $ is nontrivial so all of its elements barring the identity have order $ p $, pick two elements $g_1, g_2 $ in $ G $ with order $ p $ such that $ (g_1) \cap (g_2) = \{e\} $. (Any nontrivial element $ g_2 \notin (g_1) $ will work. To see this, assume that the intersection is nontrivial, then $g_2^a \in (g_1)$ for some $ a \neq 0 $. Now, $ a $ has an inverse mod p, say $ b $, so that $(g_2)^{ab} = g_2 \in (g_1)$, contradicting our initial choice of $g_2$.) Now, assume that $ g_1^a g_2^b = g_1^c g_2^d $, this implies that $ g_1^{a-c} = g_2^{b-d} $. However, the intersection of the cyclic groups was the identity, so we have $ a=c $ and $ b=d $. This implies that as $ a $ and $ b $ run over the congruency classes mod p, $g_1^a g_2^b $ takes $ p^2 $ distinct values, so that it runs over the entire group. To show isomorphism, we need to show that $(g_1^a g_2^b)(g_1^c g_2^d) = g_1^{a+c} g_2^{b+d}$, which follows from the result that groups of order $ p^2 $ are abelian. This implies that $ G \cong (g_1) \times (g_2) $, showing the result.