# Intuition - Quotient Group of Direct Products - Fraleigh ch. 15

Proof on p. 3 and 4 . For the case $n = 2$.

Define $h: A_1 \times A_2 \rightarrow \dfrac{A_{1}} {B_{1}} \times \dfrac {A_{2}} {B_{2}}$ as $h(a_1,a_2) = (a_1B_1, a_2B_2)$.
Apply Fundamental Homomorphism Theorem. The main result now follows by induction on n.

Fixed (1.) For Fundamental Homomorphism Theorem to operate, $\ker h := \{ \, (a_1,a_2) : h(a_1,a_2) = \color{brown}{id(\text{image of H})} \}$ needs to be $B_1 \times B_2$.

Let $id(A_n)$ = identity element of $A_n$ to reduce typing. I need to calculate $(a_1,a_2) \in \ker h$ $\iff h(a_1,a_2) = (a_1B_1, a_2B_2) \qquad = \color{brown}{( \, id(\frac{A_1}{B_1}), id(\frac{A_2}{B_2}) \,) = (idB_1, idB_2) = (B_1, B_2) }$. $\implies a_1B_1 = B_1, a_2B_2 = B_2 \iff a_1 \in B_1, a_2 \in B_2 \iff (a_1,a_2) \in B_1 \times B_2.$
Hence $\ker h = B_1 \times B_2$.

(2.) How do you envisage and envision $h(a_1,a_2) = (a_1B_1, a_2B_2)$ as the homomorphism?

(3.) What's the intuition? Right now I only understand this as a bunch of algebra.

In general, if $N$ is a normal subgroup of the group $G$, then there is always a homomorphism from $G$ onto (it is surjective) $G/N$, namely $\pi: G \rightarrow G/N$, defined by $\pi(g)=gN$ (also written as $\bar g$). Here $G/N$ is the set of cosets of $N$ in $G$, and this set of cosets can be endowed with a natural group structure: $\bar g\cdot \bar h=\overline{g*h}$, where $*$ is the group multiplication in $G$. And guess what, this is all well-defined because $N$ is normal. There is no dependence on coset representatives. Moreover, $ker(\pi)=N$.

So what you are facing is just an extension of this principle to a direct product of groups $A_i$ with each a normal subgroup $B_i$. The direct product of these $B_i$s is again a normal subgroup and the $\pi$ is treated per coordinate, so to speak.

This part is not correct, you wrote: "Let $id(A_n)$ = identity element of $A_n$ to reduce typing. I need to calculate $\ker h \iff h(a_1,a_2) = (a_1B_1, a_2B_2) = ( id(A_1), id(A_2))$". No! You have to show that $h(a_1,a_2) = (a_1B_1, a_2B_2) = (\, id(A_1/B_1), id(A_2/B_2))$, so as cosets $a_1B_1=B_1$ and $a_2B_2=B_2$. This is equivalent to $a_1 \in B_1$ and $a_2 \in B_2$. And I am sure you can take it from here now.

Hope this helps.

• Thanks a lot. Upvoted. I fixed my (1). What does the bar over your variables mean? – Group Theory Feb 24 '14 at 7:26
• The bar indicates the group elements seen in the factor groups (so really cosets). It is very common to write that with overbars. – Nicky Hekster Feb 24 '14 at 23:50

Taking quotient groups is, intuitively, "modding things out," i.e. setting things equal to zero and then living with the consequences (if $2=0$ then $1+1=0$ and $4=0$ and $1=3$ etc. for example).

This is why the congruence relation approach to quotient groups is discussed. Say $S$ is some subset of some group $G$. If you set everything in $S$ equal to the identity of $G$, there will be consequences to this. In particular, the product and inverses of anything in $S$ must also be the identity (since performing multiplication and inversion with just the identity element will yield nothing except for the identity element). Thus, if you mod out by $S$, you end up also modding out the subgroup that $S$ generates (the subgroup it generates is precisely the elements formed out of repeated multiplication and inversion of the elements of $S$). Moreover, conjugating an element of $S$ by anything from $G$ will also yield the identity, since conjugating the identity element yields the identity element. We end up not just modding out by $\langle S\rangle$, but by the conjugates of all of the elements therein, and then all of their products and inverses, and so on. Therefore, modding out by $S$ is the same as modding out by the normal subgroup generated by $S$.

In my opinion, this is an extremely important intuition in group theory.

Modding $A_1\times\cdots\times A_n$ out by $B_1\times\cdots\times B_n$ amounts to setting all elements of $B_1$ equal to the identity in the first coordinate $A_1$, etc. etc., and setting all elements of $B_n$ equal to the identity in the last coordinate $A_n$, all independently since one can form elements in $B_1\times\cdots B_n$ by choosing coordinates from the relevant $B_k$ all independently. This independence tells us that the product of quotients $A_1/B_1\times\cdots A_n/B_n$ achieves the exact same effect: in the $k$th coordinate we have the elements of $A_k$ except the elements of $B_k$ have been set equal to the identity. Since the whole setup is the same throughout (still have elements of $A_k$ for $k=1,\cdots,n$ put into coordinates), only things are being modded out, it's clear what the isomorphism $(\prod A_i)/(\prod B_k)\to\prod A_i/B_i$ should be: we simply take the residue $(a_1,\cdots,a_n)\bmod B_1\times\cdots B_n$ to $(a_1\bmod B_1,\cdots, a_n\bmod B_n)$.

• Thanks a lot. Upvoted. Can you please flesh out $4 =0, 1 = 3$ in your first paragraph? Aren't these false? Did you use this modular arithmetic? Why are you authorized to set things to the identity element? I don't understand how you can just change things in a group. Can you please notify me in your answer, and not in comments? – Group Theory Feb 24 '14 at 7:15
• @Frank I already explained in my answer. $4=0$ follows from $2=0$. Sure you can set things equal to other things. You end up creating something different from the original setting obviously. That's part of the congruence relation approach to quotients - which, again, I mentioned in my answer. Please reflect on it. – anon Feb 24 '14 at 7:22
• I'll reflect and get back to you. – Group Theory Feb 24 '14 at 7:26