# Group A is a quotient of Group B ?

• Newtime

#### Newtime

"Group A is a quotient of Group B"?

What does this phrase mean? I see it every now and again and can't figure it out. Are they say group B is the homomorphic image of group A? I'm familiar qith quotient groups, but with only groups A and B named, how would we know which quotient group of A is equal to B? This is what makes me think the former is correct. Also, I saw this used in the sense "group G is a quotient of the Free group on n letters." Since every group is the homomorphic image of a free group, I again thought my first idea was correct. Am I way off?

Nope, you're right, it should mean there is a surjective homomorphism from B to A. Usually if the homomorphism is not specified, there is a somewhat canonical one that is understood.

What does this phrase mean? I see it every now and again and can't figure it out. Are they say group B is the homomorphic image of group A? I'm familiar qith quotient groups, but with only groups A and B named, how would we know which quotient group of A is equal to B? This is what makes me think the former is correct. Also, I saw this used in the sense "group G is a quotient of the Free group on n letters." Since every group is the homomorphic image of a free group, I again thought my first idea was correct. Am I way off?

Could you given an example (link) of where this phrase is used to give us some context?

Off the top of my head saying:

"B is the homomorphic image of A"
is equivalent to saying there exists a homomorphism from A to B.

In higher level abstract algebra one may define algebraic object in terms of free objects modulo identities.

Thus for example the complex numbers can be defined as the set of polynomials in the variable i, modulo the identity i2 +1 (=0).

Another example is the tensor algebra on vector spaces. The tensor product is essentially the free (linear) product.

You can define a free group on say 2 letters by assuming the letters, a, b are generators of a group and all expressions which are not identical (modulo associativity and inverse property) are distinct.

Hence the free group generated by a and b contain distinct elements defined by all products of powers (+ and -) of a and b. Example: a-2b3a14b9
No rearrangement is possible without some identity relating ab to ba.

You can then invoke an identity such as ab = b-1a to define a smaller group.

Why do it this way? Well when comparing groups it is often easier to show the identities defining them are equivalent than to directly construct an isomorphism. Also if one set of identities implies another set (but not vis versa) that tells us there is a homomorphism (in a direction dual to implication). In general one can study algebraic objects by studying the identities. There is a category relationship and a functor between sets of identities and the objects they define from free objects.

It means the obvious thing that there exists a normal subgroup C of B such that A is isomorphic to B/C (B/C is a quotient of B).

If I understand you correctly, then your interpretations are equivalent (though I assume you meant that A is the homomorphic image of B).

If A is the homomorphic image of a group B, then there exists a surjective group homomorphism $\varphi : B \to A$ so by the first isomorphism theorem A is isomorphic to $B/\ker \varphi$.

On the other hand if there exists some normal subgroup C of B such that A is isomorphic to B/C, then there exists an isomorphism $\varphi : B/C \to A$ and in general we have a canonical projection $\pi : B \to B/C$. The composite $\varphi \circ \pi : B \to A$ is then a surjective group homomorphism (it's surjective since both $\pi$ and $\varphi$ are surjective).

You're right that often you don't know the exact quotient group, but sometimes it's enough to know that it's a quotient group.

Also, I saw this used in the sense "group G is a quotient of the Free group on n letters." Since every group is the homomorphic image of a free group, I again thought my first idea was correct.
Every group is the homomorphic image of a free group, but not every group is the homomorphic image of a free group on n letters. For instance the free group F on 3 letters is NOT the homomorphic image of a free group on 2 letters. Let S be an infinite set and let F be the free group on S. Then F is not the homomorphic image of any free group on a finite number of letters.

thanks for all the replies! I'm certain i understand it now - it seems i underestimated the importance of the first isomorphism theorem - and to clarify, the title should read : "Group B is a quotient of Group A" - i made a typo. also, there were several cases where this came up but i can't seem to find any of the best examples but here is one:

Let G be a group given as a quotient f: F(S)--->G of the free group on the set S.

Let G be a group given as a quotient f: F(S)--->G of the free group on the set S.

In this case the way to interpret it is that G is isomorphic to the quotient F(S)/ker f (assuming of course that f is surjective). Sometimes you won't be given the quotient directly, but you will instead be given an object from which you can construct the quotient.

In this case the way to interpret it is that G is isomorphic to the quotient F(S)/ker f (assuming of course that f is surjective). Sometimes you won't be given the quotient directly, but you will instead be given an object from which you can construct the quotient.

yes, it makes sense now. but this brings up another question: why say what they said instead of just saying G is the homomorphic image of F(S)? are the two compeltely interchangeable? I can see now how they are equivalent but perhaps saying one instead of the other implies more information?

Yes they are completely interchangeable (the two constructions in my first post shows how to go from one to the other). What you choose to say is a matter of taste, custom and context.

Yes they are completely interchangeable (the two constructions in my first post shows how to go from one to the other). What you choose to say is a matter of taste, custom and context.

makes sense...and thanks again, it's all quite clear now.