Prove Group of Order 15 is Cyclic

[xmcestmoi]

Please help with this question:

let ∣G∣=15. If G has only one subgroup of order 3 and only one of order 5, prove that G is cyclic.

Thank you!

 

[xmcestmoi]

I know that any group with a prime order is cyclic.
But what about a group with an order pq, where p and q are distinct prime numbers?
Is this group also cyclic?

 

[rs1n]

I know that any group with a prime order is cyclic.
But what about a group with an order pq, where p and q are distinct prime numbers?
Is this group also cyclic?

Here are some tips:

1. Write down precisely what it means for a group to be cyclic (i.e. the definition of a cyclic group). Make sure you understand this definition first.

2. If H is the only subgroup of order p and K is the only subgroup of order q, what can you say about the size of the set H\cup K? And then how would the size of this set compare to |G|?

3. What can you say about the order of an element in H? in K? and the order of an element not in H\cup K?

 

[xmcestmoi]

Thank you!
1. for a group G of order pq to be cyclic, it needs to be generated by an element with order pq.

2. If H is the only subgroup of order p and K is the only subgroup of order q , then the size of the set HUK is P+q-1, because the identity element of G is the only element in both H and K.
This means there is one element x in G, but not in H, and not in K, either.

3. the order of an element in H will have to be 1, or p.
the order of an element in K needs to be 1, or q.
so the element x must have an order pq, x generates G.

Your hints help a lot!

But is this reasoning (my answer to your hint questions) complete? or is there still something missing?

 

[Dick]

You are right that the order of HUK must be 3+5-1=7. That means there must be an element x of G that is in neither H nor K. What's the order of x? Why can't it be 3 or 5?

 

[xmcestmoi]

I suspect that

if x has order 3, then it generates H
and similarly, if x has order 5, then it generates K
in either case, x will be in H or K
contracting previous fact that x of G is in neither H nor K

 

[Dick]

I suspect that

if x has order 3, then it generates H
and similarly, if x has order 5, then it generates K
in either case, x will be in H or K
contracting previous fact that x of G is in neither H nor K

I'd be a little more specific on WHY if x has order 3 it generates H. All you know from the fact x has order 3 is that it generates SOME subgroup of order 3.

 

[xmcestmoi]

isnt it true that any element in G, say an element x will generate a cyclic group with order = the order of x ?

and then going back to the original question , which says "only one subgroup of order 3"

so <x> is "the" subgroup H

 

[Dick]

isnt it true that any element in G, say an element x will generate a cyclic group with order = the order of x ?

and then going back to the original question , which says "only one subgroup of order 3"

so <x> is "the" subgroup H

Right. I just wanted you to state the "only one subgroup" premise. Otherwise it isn't necessarily true. There are examples of noncyclic groups of order pq. But not if there is a unique subgroup of order p and order q.

 

[xmcestmoi]

Thank you :)

I did not occur to me before

 

[xmcestmoi]

another question concerning a group of order pq

if the order of a group G is pq, where p,q are distinct prime numbers,

then G has two normal subgroups, one has order p, and the other has order q.

Is this statement true?

 

[Dick]

another question concerning a group of order pq

if the order of a group G is pq, where p,q are distinct prime numbers,

then G has two normal subgroups, one has order p, and the other has order q.

Is this statement true?

No. Why would you think so? If that were true every group of order pq would be cyclic. Isn't that basically what we just proved. It's not true every group of order pq is cyclic.

 

[xmcestmoi]

No. Why would you think so? If that were true every group of order pq would be cyclic. Isn't that basically what we just proved. It's not true every group of order pq is cyclic.

I was just confused by a problem :(

But I get your point.Thank you!