Solve Induction Problem: Show Group G of Order p^n is Cyclic

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Homework Help Overview

The problem involves a group G of order p^n, where p is a prime number, and requires showing that G is cyclic given that it has exactly one subgroup for each divisor of its order.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the inductive approach to proving the claim, questioning the properties of the center of the group Z(G) and its implications for the structure of G.
  • Some participants raise questions about the nature of divisors of p^n and the existence of subgroups, particularly regarding the trivial subgroup.
  • There is a discussion about the necessity of Z(G) being non-trivial in the context of p-groups and the implications for the quotient group G/Z(G).

Discussion Status

The discussion is ongoing, with participants exploring various aspects of group theory related to the problem. Some guidance has been offered regarding the use of the inductive hypothesis and the properties of the center of the group, but no consensus has been reached.

Contextual Notes

Participants note that the group G is of order p^n, which imposes certain constraints on the structure of its subgroups and the center Z(G). There is also mention of the Fundamental Theorem of Finite Abelian Groups as a potential tool for analysis.

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Homework Statement


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.


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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.
 
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What are the divisors of p^{n}?
 
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
 
Every group has 2 subgroups, namely \{e\} and G. So, there is always a subgroup with 1 element.
 
{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.
 
I'm sorry. Maybe it's trivial, but I am unfamiliar with the notation Z(G). What do you mean by it?
 
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).
 
You're probably supposed to use the inductive hypothesis on the quotient G/Z(G)
 

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