What is the minimal subgroup that contains two arbitrary subgroups?

[farleyknight]

Just getting into group theory, so don't be surprised if this doesn't make any sense. Also, since I'm a novice, I'm assuming this kind of group has already been named, although I can't find an example in any of my books or on Google. Does anyone recognize this? I'd like to study it further..

Let <a> = A and <b> = B be two cyclic groups with |a| = n and |b| = m. Since <a> is a subgroup of S_n and <b> is a subgroup of S_m then there must be some larger group for S_{n+m} so that both <a> and <b> are subgroups. So then define the subgroup of S_{n+m} which are all possible products of a and b.

I guess another way to describe it would be the minimal union of two disjoint groups <a>, <b>.

Any ideas?

 

[yyat]

Let <a> = A and <b> = B be two cyclic groups with |a| = n and |b| = m. Since <a> is a subgroup of S_n and <b> is a subgroup of S_m then there must be some larger group for S_{n+m} so that both <a> and <b> are subgroups.

It is a bit ambiguous to say a certain cyclic group of order n is a subgroup of $$S_n$$, because $$S_n$$ has many subgroups which are cyclic of order n (for larger n).
However, there certainly exists a group which has A and B as subgroups: the product group $$A\times B$$, but this is not necessarily the smallest group containing A and B as subgroups.

 

[ThirstyDog]

I the approach to find the minimal group would be something like:
Let d = gcd(n,m).

Define the group <c> where c has order mn/d. In this group
$$a = c^{m\d} and b = c^{n\d}$$

To check this is minimal check the order of ab this will tell you the minimal order of the group which has a and b in it.

Note: I have assumed the minimal group will be abelian - I can't think of an obviously simple explanation of why this must be. This is a point to check.

 

[Hurkyl]

farleyknight: the general construction you're trying to think of is called a coproduct. In the specific case of groups, it is apparently called the free product.

p.s. your example doesn't actually achieve this goal; I'm fairly sure the free product of two (nontrivial) cyclic groups is infinite.

 

[John Creighto]

p.s. your example doesn't actually achieve this goal; I'm fairly sure the free product of two (nontrivial) cyclic groups is infinite.

Wikipedia seems to agree with you. I'm not sure why this is the case though. I must be missing something in the definition though because if the elements commute then I would think the group would be finite. I must be thinking of the wrong type of product. I'll look at it again later.

 

[Hurkyl]

if the elements commute

The operative word is "if"... If you take the free product and impose additional relations that say that elements of the first group commute with elements of the second group, then you do indeed get the Cartesian product.


(Of course, if you're doing Abelian group theory, then those relations wouldn't be additional; in that theory, the free product and the Cartesian product are isomorphic)

 

[ThirstyDog]

Let <a> = A and <b> = B be two cyclic groups with |a| = n and |b| = m.
...
I guess another way to describe it would be the minimal union of two disjoint groups <a>, <b>.

Is the question what is the smallest group containing a and b (in which case the Cartesian product) or the smallest group that contains isomorphic copies of A and B (in which case I think it is what I wrote before)?

 

[MathematicalPhysicist]

Well, the smallest group which contains both A and B, is their join, AvB.

 

[ThirstyDog]

I know this thread is pretty much dead but...

The reason that <c> where c has order mn/d (where d = gcd(m,n)) is the smallest group containing isomorphic copies of <a> and <b> (where |a|=n, |b|=m) is that the order c is precisely the lowest common multiple of n and m.

 

[matt grime]

I'm fairly sure the free product of two (nontrivial) cyclic groups is infinite.

The free product of any two (non-trivial) groups is infinite.

 

[matt grime]

Well, the smallest group which contains both A and B, is their join, AvB.

What's the definition of join? And is it purely an isomorphism independent notion? There is a difference between equals, and isomorphic to bear in mind, as ever.

If I were to say that G is a group of permutations of {1,2,3} generated by the transposition (12), and H were the group generated by (123), you could argue that the smallest group containg G and H is the full permutation group of {1,2,3}, but this is "remembering" how I defined G and H. There is another group of order 6 containing groups isomorphic to G and H (but not equal).

 

[MathematicalPhysicist]

What's the definition of join? And is it purely an isomorphism independent notion? There is a difference between equals, and isomorphic to bear in mind, as ever.

The least set which contains A and B.

 

[matt grime]

Well, that isn't even a group.

But then "the smallest group that contains two arbitrary groups" doesn't exist: is the smallest group that contains (something isomorphic to) C_2 and C_3 either C_6 or S_3?It does make sense to talk of the smallest subgroup of G that contains two subgroups H and K, though.