# Thread: induction on p-groups

1. ## induction on p-groups

Let $p$ be a prime and let $G$ be a group of order $p^\alpha$. Prove that $G$ has a subgroup of order $p^\beta$, for every $\beta$ with $0<=\beta<=\alpha$.

Hint: induction on $\alpha$ and use the theorem: If $p$ is a prime and $G$ is a group of prime power order $p^\alpha$ for some $\alpha>=1$, then $G$ has a nontrivial center.

2. Originally Posted by dori1123
Let $p$ be a prime and let $G$ be a group of order $p^\alpha$. Prove that $G$ has a subgroup of order $p^\beta$, for every $\beta$ with $0<=\beta<=\alpha$.

Hint: induction on $\alpha$ and use the theorem: If $p$ is a prime and $G$ is a group of prime power order $p^\alpha$ for some $\alpha>=1$, then $G$ has a nontrivial center.
if $\alpha = 0 \ \text{or} \ 1,$ then there's nothing to prove. so let $\alpha > 1$ and suppose that the claim is true for all groups of order $p^{\beta}$ with $0 \leq \beta < \alpha.$ let G be a group of order $p^{\alpha}$ and let $0 \leq \gamma \leq \alpha.$ we want

to show that G has a subgroup of order $p^{\gamma}.$ it's trivial for $\gamma = 0.$ so we'll assume that $1 \leq \gamma \leq \alpha.$ we have to consider two cases:

Case 1: G is abelian. in this case choose $x \in G$ with $o(x)=p.$ since G is abelian, < x > is normal in G. now G/< x > is a group of order $p^{\alpha - 1}.$ so by induction hypothesis it has a subgroup H/<x>

of order $p^{\gamma -1}.$ therefore: $|H|=p^{\gamma}$ and we're done in this case.

Case 2: G is non-abelian. let Z(G) be the center of G and $|Z(G)|=p^{\delta}.$ then: $1 \leq \delta < \alpha.$ if $\gamma \leq \delta,$ then by induction hypothesis Z(G) has a subgroup of order $p^{\gamma}$ and we're done. if $\gamma > \delta,$ then

consider G/Z(G), which is a group of order $p^{\alpha - \delta}.$ by induction G/Z(G) has a subgroup K/Z(G) of order $p^{\gamma - \delta}.$ hence: $|K|=p^{\gamma}. \ \ \Box$