# Help Understanding Quotient Groups? (Dummit and Foote)

1. Jun 4, 2013

### robertjordan

The definition given is...

"Let $\phi: G \rightarrow H$ be a homomorphism with kernel $K$. The quotient group $G/K$ is the group whose elements are the fibers (sets of elements projecting to single elements of H) with group operation defined above: namely if $X$ is the fiber above $a$ and $Y$ is the fiber above b then the product of $X$ and $Y$ is defined to be the fiber above the product $ab$."

But what if we have a homomorphism $\alpha: G \rightarrow A$ and a homomorphism $\beta: G \rightarrow B$ that both have kernel $K$?

Wouldn't this mean $G/K$ is not unique because $\alpha: G \rightarrow A$ requires $G/K$ to be fibers consisting of elements in $A$ whereas $\beta: G \rightarrow B$ requires $G/K$ to be fibers consiting of elements in $B$?

I'm confused. :(

2. Jun 4, 2013

### micromass

Staff Emeritus
That's a pretty horrible definition.

But anyway, you can show that $G/K$ as defined by the map $\alpha$ will be isomorphic to $G/K$ as defined by $\beta$.

3. Jun 7, 2013

### robertjordan

I left out one part of the definition. It should say

"Let ϕ:G→H be a homomorphism with kernel K. The quotient group G/K is the group whose elements are the fibers of ϕ (sets of elements projecting to single elements of H) with group operation defined above: namely if X is the fiber above a and Y is the fiber above b then the product of X and Y is defined to be the fiber above the product ab."

WANT TO SHOW: "If $\alpha: G \rightarrow A$ and $\beta \rightarrow B$ have the same kernel, K, then the quotient group whose elements are the fibers of $\alpha$ is isomorphic to the quotient group whose elements are the fibers of $\beta$."

... I don't know how to show this. :(
This would mean $|image(\alpha)|=|image(\beta)|$, but I don't know how to show that either.

Could I have a push in the right direction?

EDIT: I'm working on Office_Shredder's response right now...

Last edited: Jun 7, 2013
4. Jun 7, 2013

### Office_Shredder

Staff Emeritus
The more normal definition (haha pun intended as will become clear) is typically this:

If K is a subgroup of G with the following property: for all $g\in K$ and $h\in G$, $h^{-1} g h \in K$ we call K a normal subgroup.

G/K is then defined as the set
$$\{ hK\ |\ h\in G \}$$
is the set of all cosets of G, with the multiplication operation (hK is the set of all elements hg with g in K)
$$\left(hK \right) \cdot \left( gK \right) = \left( (hg) K \right)$$
K being normal is exactly the property required for this to be a group (exercise: prove this)

Cool facts: The kernel of a homomorphism is always a normal subgroup (easy exercise to prove). Also, with the above construction every normal subgroup is also the kernel of a group homomorphism, in particular the projection $p: G \to G/K$ with $p(g) = gK$ (the kernel is precisely K)

The (first?) isomorphism theorem of group theory says that if $\phi:G\to H$ is a group homomorphism. then $G/(ker(\phi)) = \phi(G)$ i.e. the groups on the left and right are isomorphic, and phi is an isomorphism between them - phi is defined on G/(ker(phi)) by $\phi( h ker(\phi)) = \phi(h)$ (you can check this is well defined and still defines a group homomorphism)

Now given K we could have picked an arbitrary $\phi$ with kernel K, $\phi: G\ to H$ and we see that the fiber definition in the original post gives a group isomorphic to the definition of G/K at the top of this post. Dummit and Foote decided on a different path for defining quotient groups but I think it makes it a lot harder to do calculations than this one (and is a lot less intuitive)

5. Jun 7, 2013

### robertjordan

I don't see why K needs to be normal?

In order for $(G/K,\cdot)$ to be a group we need:

i) $g_1K\cdot(g_2K\cdot{g_3K})=(g_1K\cdot{g_2K})\cdot{g_3K}$.
ii) $\exists{e\in{\frac{G}{K}}}$ s.t. for all g in $\frac{G}{K}$, $e\cdot{gK}=gK$.
iii) for every $gK\in{\frac{G}{K}}$ we need there to exist a $(gK)^{-1}$ s.t. $(gK)(gK)^{-1}=e$.
iv) we need the set to be closed under the operation $\cdot$.

i) is satisfied because $g_1K\cdot(g_2K\cdot{g_3K})=g_1K\cdot((g_2g_3)K)=(g_1(g_2g_3))K=((g_1g_2)g_3)K=(g_1g_2)K\cdot{g_3K}=(g_1K\cdot{g_2K})\cdot{g_3K}$.

ii) is satisfied because $e_g{K}\cdot{gK}=(e_gg)K=gK$.

iii) is satisfied because $g^{-1}K\cdot{gK}=(g^{-1}g)K=e_gK$, which is the identity of G/K.

iv) the set is closed under the operation $\cdot$ because if $g_1K$ and $g_2K$ are any two arbitrary elements in the set, then since $g_1K\cdot{g_2K}=(g_1g_2)K$, and since $(g_1g_2)\in{G}$, we know $(g_1g_2)K$ is in the set "G/K".

So why do we need K to be normal?

Last edited: Jun 7, 2013
6. Jun 7, 2013

### micromass

Staff Emeritus
Because you want that if $aK = a^\prime K$ and $bK = b^\prime K$, then $(aK)(bK) = (a^\prime K) (b^\prime K)$. This is precisely guaranteed if $K$ is normal.

7. Jun 7, 2013

### robertjordan

I don't understand, why can't we just say:

If $aK=a'K$ and $bK=b'K$, then $(aK)(bK)=(a'K)(bK)=(a'K)(b'K)$.

?

8. Jun 7, 2013

### Office_Shredder

Staff Emeritus

Pick some g, and h and h' such that hK = h'K. Is (hg)K = (h'g)K?
This is the same as h(gK) = h'(gK) (this part conforms to your intuition for how these things work) which is NOT immediately implied by hK = h'K (because gK has different elements than K does)