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!


We can do part a) by direct application of the class equation. Assume G doesn't have a subgroup of order p^k. Note that Cl(ai) is not 1 for all i(as if it was 1 then ai would have just been a part of Z(G)) also Cl(ai) is not equal to q(as if it were equal we would get a subgp of order p^k) therefore as |G| is divisible by p and Cl(ai) is also divisible by p for all i. Z(G) is divisible by p.

2nd part is just because of Cauchy's theorem. For every prime dividing a group G there exists an element x belonging to G of order p. (a proof of this can be found anywhere on the internet)

  • $\begingroup$ 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 $p$? $\endgroup$ – Kamil Nov 3 '16 at 17:33
  • $\begingroup$ |Cl(ai)| = |G:Cg(ai)| if Cl(ai)=q => |Cg(ai)| =p^k this would give a contradiction. Also now |G:Cg(ai)| divides order of G. It can't be 1 or q. Therefore it has to have some factor of p in it $\endgroup$ – user385655 Nov 3 '16 at 17:41

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