Groups of Prime Power Order: Must There Be an Element of Order p?

[Kate2010]

Homework Statement

Any help with this question would be great:

G is a group such that |G| = p^k, p is prime and k is a positive integer. Show that G must have an element of order p.

The hint is to consider a non-trivial subgroup of minimal order.

Homework Equations

Lagrange

The Attempt at a Solution

Can I use Lagrange to say that there must exist a subgroup H of order p^m with m<k? Or even a subgroup H of order p? Although I'm not sure how to justify this. Then if H was cyclic then there must be an element of order p? Again, this is just a guess.

 

[Tedjn]

You are right to suspect that Lagrange's theorem cannot be used to justify that there exist subgroups of certain sizes. It can only be used directly to justify that there do not exist subgroups of certain sizes. Finding a (cyclic) subgroup H with p elements is enough, if you can do so, because the generator of H would have order p. (Why?) How can you find cyclic subgroups in G?

 

[Kate2010]

I don't know?

All I know is that G contains p^k elements. I'm a bit confused about how to deduce anything else about the group.

 

[Tedjn]

Right, since we don't know anything else about this group, let's look at the elements of it. What order can these elements have?

 

[Kate2010]

The order of the elements must divide p^k

So they could be order p, and then p^t where t divides k.

 

[Tedjn]

So if there is an element of order p, then we are done. Suppose not. Then, every (nonidentity) element has order p², p³, ..., or p^k. Now, remember that a cyclic group is generated by one element; a cyclic subgroup is generated by one element of this group. What do these cyclic subgroups look like?

 

[Kate2010]

<p> = {p^k : k is an integer}

This looks like the list you have just given.

 

[Martin Rattigan]

The order of the elements must divide p^k

So they could be order p, and then p^t where t divides k.

Care! You only need t<=k.
1,3,9,27 all divide 27 for instance.
That is 3^t with t=0,1,2,3 all divide 3^k with k=3.

 

[Tedjn]

Yes, that's the definition of a cyclic subgroup for an element p. Let's use the letter g to denote an arbitrary element from now on to avoid confusion with the prime p. <g> is defined to be all the powers of g. But since g is an element of a finite group G, the cyclic subgroup generated by g must have finite order. In particular, the order is the order of g, which is the generating element.

By assumption g has order p^m for some 1 < m <= k, so <g> has order p^m. Is there a natural nontrivial subgroup of <g> that has order less than p^m?

EDIT: I need to step our for a few hours. If you haven't figured it out, I'm sure someone else will pick up from where I left off. What I am trying to illustrate is the way I've come to reason about groups when I see such a problem. Thinking about orders of group elements and orders of groups. Thinking about what if there were no element of order p; what contradiction can arise? What particular subgroups are easy to identify, e.g. cyclic subgroups? Granted I am taking algebra this semester as well, and my viewpoint is not very advanced, but I hope that this helps a bit more than just giving you the answer.

 

[Kate2010]

So is <g> = {e, g, g^2, ..., g^(p^m -1)}?

I'm still a bit stumped as to how to find a subgroup.

 

[Tedjn]

Yes, that lists explicitly the elements of <g>. We know <g> is a cyclic subgroup of G with p^m elements. What possibly subgroups of this cyclic group are there? Such a nontrivial subgroup must include at least one nonidentity element. It must then contain all powers of that element by closure. Is there a nice choice for this first element that leads to a smaller subgroup? What sizes can you make these subgroups?

 

[Kate2010]

What if we chose <g^p> ?

I'm still not too sure, but ready to quite for the night now too. Thank you for all your help :)

 

[Tedjn]

You are almost there. What is the order of <g^p>? What does that tell you about the order of the element g^p? Is there another element that is better to use?

 

[Kate2010]

Is the order of <g^p> m? Then the order of g^p is also m?

 

[Tedjn]

Careful there. Is (g^p)^m equal to g^pm?