# induction problem

#### wutang

[FONT=&quot]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. I am suppose to use induction to solve this problem, but I am getting no were on it. Do I need to use a Sylow Theorem in order to get it were I can use induction? I need help!!!
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#### TheArtofSymmetry

[FONT=&quot]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. I am suppose to use induction to solve this problem, but I am getting no were on it. Do I need to use a Sylow Theorem in order to get it were I can use induction? I need help!!![/FONT]
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. Now, by inductive hypothesis, G/Z(G) is cyclic. Thus G is abelian.

By the classification theorem of finitely generated abelian groups and inducitive hypothesis, $$\displaystyle G \cong \mathbb{Z}_{p^{n-1}}$$$$\displaystyle \times \mathbb{Z}_p$$ or $$\displaystyle G \cong Z_{p^n}$$. The former is contradictory to the inductive hypothesis because there are more than one subgroup of order p in G. Thus $$\displaystyle G \cong Z_{p^n}$$ and G is cyclic.

wutang

#### wutang

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.
So Z(G) is either p, p^2,...,p^k by lagrange. If Z(G) is p or P^2 then G/Z(G) would be non-trivial cyclic, which cannot happen, so Z(G) must be p^k right? Besides that your explanation makes perfect sense and has really helped me on this problem.

#### TheArtofSymmetry

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.
So Z(G) is either p, p^2,...,p^k by lagrange. If Z(G) is p or P^2 then G/Z(G) would be non-trivial cyclic, which cannot happen, so Z(G) must be p^k right? Besides that your explanation makes perfect sense and has really helped me on this problem.
Let G be a group of order p^n satisfying the given conditions. Once we know that Z(G) is a non-trivial p-subgroup of G, we can derive that G/Z(G) is a p-group having an order less than or equal to p^{n-1}. We assumed that the claim is true for all orders of p^k where k<n (inductive hypothesis).
By inductive hypothesis, G/Z(G) is cyclic and we conclude that G is abelian.

wutang

#### wutang

If |G|=81=3^4, so Z(G) is non-trivial, so Z(G)=3,3^2, or 3^4. I am not seeing how you are eliminating 3 and 3^2 in this case because in order for G/Z(G) to be cyclic, |Z(G)|=|G|. I am totally lost, save me please!

#### TheArtofSymmetry

If |G|=81=3^4, so Z(G) is non-trivial, so Z(G)=3,3^2, or 3^4. I am not seeing how you are eliminating 3 and 3^2 in this case because in order for G/Z(G) to be cyclic, |Z(G)|=|G|. I am totally lost, save me please!
If G is abelian, then Z(G) is the whole G, which follows G/Z(G) is a trivial cylic group. We ususally use

G/Z(G) cyclic --> G abelian.

Actually if G/Z(G) is cyclic, it always results in G/Z(G)={e}. Anyhow it is a very useful theorem to derive G is abelian.

What was the inductive hypothesis?

The inductive basis where n=1 trivially holds because the group of order p is cyclic. The inductive hypothesis says that the groups of order p^k where k<n are cyclic groups.

We don't need the exact order of Z(G). We only use Z(G) is a non-trivial p-subgroup of G. Then G/Z(G) is the group of order p^i, where i<n. Again, by the inductive hypothesis, G/Z(G) is cyclic and we conclude that G is abelian. Now we are half-done. The remaining part is a simple procedure by using the classification theorem of finitely generated abelian groups.

wutang

#### wutang

We don't need the exact order of Z(G). We only use Z(G) is a non-trivial p-subgroup of G. Then G/Z(G) is the group of order p^i, where i<n. Again, by the inductive hypothesis, G/Z(G) is cyclic and we conclude that G is abelian.
Since |G/Z(G)|=p^i with i<n, then we know that |G/Z(G)| is cyclic since we have already show that all groups of order p^i with i less than n are cyclic. Thus by the G/Z(G) thereom, G is abelian. Is this what you are saying because it is starting to make a lot of sense to me finally!!!

#### TheArtofSymmetry

We don't need the exact order of Z(G). We only use Z(G) is a non-trivial p-subgroup of G. Then G/Z(G) is the group of order p^i, where i<n. Again, by the inductive hypothesis, G/Z(G) is cyclic and we conclude that G is abelian.
Yeah.

Since |G/Z(G)|=p^i with i<n, then we know that |G/Z(G)| is cyclic since we have already show that all groups of order p^i with i less than n are cyclic. Thus by the G/Z(G) thereom, G is abelian. Is this what you are saying because it is starting to make a lot of sense to me finally!!!

Read the "Proof by strong induction" part (here).

I didn't show that all groups of order p^i with i less than n are cyclic. I just assumed that the claim is true for all groups of order p^i with 1<= i < n, i.e., they are cyclic and they have exactly one subgroup having an order for each divisor of p^i, respectively. It is an inductive hypothesis.

Our goal is to show that the claim is true for all finite p-group G with |G|=p^n, so the choice of n is arbitrary. That is why we used an induction.

wutang