induction problem

Apr 2010
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[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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May 2010
95
38
[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.
 
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Apr 2010
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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.
 
May 2010
95
38
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.
 
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Apr 2010
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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!
 
May 2010
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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.
 
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Apr 2010
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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!!!
 
May 2010
95
38
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.
 
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