# Group of order $p^k q$ where $p$ and $q$ are distinct primes - how to prove this property?

I'm trying to figure out this problem:

Problem: Let $G$ be a group of order $p^k q$ with $p$ and $q$ distinct primes and $k \in \mathbb{N} \setminus \left\{0\right\}$.

a) Prove: $G$ has a subgroup of order $p^k$, or $p$ divides $|Z(G)|$. (Here $Z(G)$ means the center of $G$).

b) Prove: if $p$ divides $| Z(G)|$, there there exists an $x \in Z(G)$ of order $p$.

Attempt: For a) I tried proceeding by induction.

Base step: Assume $k=1$. Then $|G| = pq$ with $p \neq q$. Case 1: if $G$ is commutative, then $Z(G) = G$ and then $p$ divides $|Z(G)|$.

Case 2: Assume $G$ is not commutative. Then the center of $G$ is trivial, i.e. $Z(G) = \left\{e_G\right\}$ (this is a theorem of group theory). So $p$ does not divide $|Z(G)|$. Then we must show that $G$ has a subgroup of order $p$. Let $H$ be a non-trivial subgroup of $G$, that is $H \neq G$ and $H \neq \left\{e\right\}$. Then by Lagrange, $|H|$ divides $pq$. So we must have either that $|H| = p$ or $|H| = q$ (is this correct? Since $H$ is non-trivial). If $|H| = p$ then we are done. If $| H | = q$, then $|G / H| = p$ and so $G/H$ is cyclic. I want to derive a contradiction from this, and so show that $H$ cannot have order $q$, but I don't know how.

Induction step: Assume the claim holds for $|G| = p^{k-1} q$. We must then show that it also holds for $|G| = p^k q$. I want to use the class equation. We have that $$p^k q = |Z(G)| + \sum_{i=m+1}^n |Cl(a_i)|$$ where $|Cl(a_1)| = \ldots = |Cl(a_m)| = 1$ and $|Cl(a_i)| > 1$ for $i > m$. Since $p$ is a divisor of the LHS, it must also divide the linear combination. But how can I conclude from this that $p$ divides $| Z(G)|$ ? I don't know how to use the induction hypothesis.

b) Assume $p$ divides $|Z(G)|$. I first assume that $G$ is abelian. Then by structure theorem for finite abelian groups, we know that $$G \cong \mathbb{Z}_{p_1^{k_1}} \oplus \ldots \oplus \mathbb{Z}_{p_r^{k_r}}.$$ This means that $$p^k q = p_1^{k_1} \ldots p_r^{k_r}.$$ So $p | (p_1^{k_1} \ldots p_r^{k_r})$ and so since $p$ is prime, $p$ must divide at least one factor. Assume that $p$ divides $p_i^{k_i}$. Since $p_i^{k_i} = p_i \ldots p_i$ with $k_i$-amount of factors, we must again have that $p$ divides $p_i$. So $p = p_i$.

Now, since every $\mathbb{Z}_{p_i^{k_i}} = \mathbb{Z}_p^{k_i}$ is abelian, we have that the center equals the whole group. Let $x_i$ be an element in this center. Then $order(x_i) | p^{k_i}$ and so $order(x_i) | p$. Hence the order of $x_i$ equals $p$ in $\mathbb{Z}_p^{k_i}$. Since $x_i$ is also in $Z(G)$, this concludes the proof?

And what about the case that $G$ is not abelian?

Help with this problem is appreciated!

• Thank you for reply. Two questions about part a). Can you explain in more detail please why we can't have $|Cl(a_i) |= q$, and and why $|Cl(a_i)|$ is divisible by $p$? – Kamil Nov 3 '16 at 17:33