# Homework Help: Induction problem

1. May 16, 2010

### tyrannosaurus

1. The problem statement, all variables and given/known data
Suppose the G is a group of order p^n, where p is prime, and G has exactly one subgroup for each divisor of p^n. Show that G is cyclic.

2. Relevant equations

3. The attempt at a solution
First, the inductive basis holds for n=1. Now for each k>=1, we assume that the claim is true for a group of order p^k where k < n. We shall show that the claim is also true for a group of order p^n.

Note that in a general case, the group of order p^n is not necessarily abelian. However, by the given inductive hypotheis, we show that the above group G of order p^n is abelian. We know that Z(G) is non-trivial for every nontrivial finite p-group and is a subgroup of G.

From this I am suppose to use the G/Z(G) theroem to show that G/Z(G) is cyclic and thus that G is abelian. However, I don't see how I this is true, for example what if |G|=p^3 and |Z(G)|=p, then G/Z(G)= p^2, which does not mean that G/Z(G) is cyclic. (If I remember right G/Z(G) has to be trivial cyclic)
From this I can use the Fundamental Theorem of Finite Abelian Groups to get the G is isomorphic to Zp+Z(p^n-1) or Zp and then conclude that it is isomorphic to Zp.

2. May 16, 2010

### Dickfore

What are the divisors of $p^{n}$?

3. May 16, 2010

### tyrannosaurus

1, p, p to m where m divides n and p^n. We know that we can't have 1 because Z(G) needs more than one lement

4. May 16, 2010

### Dickfore

Every group has 2 subgroups, namely $\{e\}$ and $G$. So, there is always a subgroup with 1 element.

5. May 16, 2010

### tyrannosaurus

{e} is always a subgroup, but in are case Z(G) cannot equal e since are group G is of order p^n, where p is a prime.

6. May 16, 2010

### Dickfore

I'm sorry. Maybe it's trivial, but I am unfamiliar with the notation $Z(G)$. What do you mean by it?

7. May 16, 2010

### tyrannosaurus

Z(G) is the center of the group G, for a to be an element of Z(G) a has to commute with all elements of G. Z(G) can be trivial (i.e just e) in some cases, but in this one it can't since |G|=p^n where p is a prime (p0groups have non-trivial centers).

8. May 17, 2010

### Office_Shredder

Staff Emeritus
You're probably supposed to use the inductive hypothesis on the quotient G/Z(G)