# Show group equivalence relation associated with normal subgroup

1. Sep 15, 2015

### jackmell

1. The problem statement, all variables and given/known data
Let $G$ be a group and $\sim$ and equivalence relation on $G$. Prove that if $\sim$ respects multiplication, then $\sim$ is the equivalence relation associated to some normal subgroup $N\trianglelefteq G$; i.e., prove there is a normal subgroup $N$ such that $x\sim y$ iff $xN =yN$.

2. Relevant equations
An equivalence relation respects multiplication if $x\sim y$ implies that $xz\sim yz$ for all $z\in G$.

3. The attempt at a solution

I can show that if the equivalence is a partition of $G$ into cosets $gH$ with $H\leq G$ (H a subgroup of G), then in order for that equivalence to respect multiplication, $H$ has to be normal in $G$:

Define the following equivalence relation on $G$:

$x\sim y \Rightarrow x\in yH\Rightarrow xH=yH,\quad H\leq G$
where $yH$ is the representative for this coset in the partitioning of $G$. Then in order for this equivalence to respect multiplication, we would need $xzH=yzH\;\forall\; z\in G$ or $\left(yz\right)^{-1} (xz) H\in H$ . That means $z^{-1}\left(y^{-1}x\right)z\in H\;\forall\;z\in G$. Now the quantity $y^{-1}x$ represents all of $H$ so that is equivalent to $z^{-1} H z\in H$ and that would mean $H$ has to be normal.

However I don't understand how I can prove that every type of partitioning of $G$ into equivalent classes that respect multiplication necessarily is associated with some normal group or perhaps can be related to this coset partitioning.

Jack

2. Sep 15, 2015

### Dick

One of your equivalence classes contains the identity. Start by showing that it is a subgroup.

3. Sep 15, 2015

### jackmell

Ok thanks Dick. I'm working on it. So if I define an equivalence relation on a group, that necessarily partitions the group into disjoint partitions and one of these partitions, say $P_i$, will have the identity element. So that if I can show $P_i\leq G$, then I can express the group in terms of cosets of this partition and then use the argument above to conclude that any equivalence partition of a group which respects the group operation is associated with a normal subgroup?

However, although I know the basic principles for showing a set, $S$ is a subgroup, (show $1\in S$, if $x\in S$, show $x^{-1}\in S$, then show for $x,y \in S$ so too is $xy\in S$ or alternatively if $|S|\neq 0$, then only need to show for all $x,y\in S$, $x y^{-1}\in S$, not really sure how to proceed to show it's a subgroup using only equivalence statements.

Let me try:

I define an equivalence relation on $G$ such that $x\sim y$ if $x$ is in the same partition as $y$. And I want this equivalence relation to respect the group (multiplicative) operation. First assume $1\in S$. Now let $x\in S$. Then $1\sim x$. So that in order to preserve the group operation, I would need to have $1(y)\sim x y$. But let $y=x^{-1}$ so that I would need $x^{-1}\sim x x^{-1}$ or $x^{-1}\sim 1$ or $x^{-1}\in S$.

Now let $x,y\neq x^{-1}\in S$, then $x\sim y$ so that in order to preserve multiplication, I would require $xz\sim yz$. So again, let $z=y^{-1}$. Then I obtain $xy^{-1}\sim 1$ so $xy^{-1}\in S$.

I don't feel too confident about this as it's new to me. Is this the way to approach this problem? I'll work on it more.

Thanks,
Jack

Last edited: Sep 15, 2015
4. Sep 15, 2015

### micromass

Staff Emeritus
You don't define anything. You are given an equivalence relation and you know that it respect the multiplication. Your proof seems correct to me, but it is worded very weirdly. For example, you say "And I want this equivalence relation to respect the group (multiplicative) operation.". You know this by assumption, so I don't get your statement. Anyway, if you rewrite your proof a bit, it should be correct. Now show that the subgroup $P_i$ is normal.

5. Sep 15, 2015

### jackmell

Ok I see that now. I'll clean it up. Thanks guys!

Jack

6. Sep 15, 2015

### jackmell

This is my cleaned-up proof:

Partition $G$ into a disjoint collection of set $G=\{S_1,S_2,\cdots,S_n\}$ and define an equivalence relation on $G$ by $x\sim y$ if $\{x,y\}\in S_i$. Claim if this equivalence is to respect multiplication, then this necessarily induces one of the sets to be a subgroup of $G$.
First, since this is a disjoint collection of sets, one has to contain the identity element. Call this set $H$. If $|H|=1$ then we are done. Assume $|H|>1$ and let $x\in H$. Then in order to preserve multiplication, we require $1y\sim xy\;\forall y\in G$. Now let $y=x^{-1}$. Then $x^{-1}\sim 1$ or $x^{-1}$ needs to be in $H$. Assume $x,y\in S$ and again require $xz\sim yz\;\forall z\in G$. Now let $z=y^{-1}$ which implies $xy^{-1}\sim 1$ so $xy^{-1}\in S$. Therefore, in order for a partition to preserve multiplication, one of the elements of the partition must be a subgroup of $G$.

If $H\leq G$, then the cosets $xH$ and $Hx$ partition $G$. The following equivalence relations can then be defined based on these partitions:
\begin{align} x\sim y \Rightarrow xH&=yH\Leftrightarrow y^{-1} x\in H \\ x\sim y \Leftarrow Hx&=Hy\Leftrightarrow yx^{-1}\in H \\ \end{align}
Claim these equivalences respect group multiplication iff $H\trianglelefteq G$. Let $x,y\in G$ and consider the left cosets (right cosets follow by symmetry). If $x\sim y$ then $xH=yH$ so that if this equivalence relation respects multiplications, must show $xz\sim yz$ or $xzH=yzH$ for $z\in G$. This is true iff $(yz)^{-1}(xz)\in H$ or $z^{-1}\left(y^{-1} x\right) z\in H$. That means $z^{-1}(y^{-1}x)z\in H\;\forall z \in G$. Now the quantity $y^{-1}x$ represents all of $H$ so that is equivalent to $z^{-1}Hz\subseteq H$ which implies $H\trianglelefteq G$.

Conversely, assume there exists $N\trianglelefteq G$. Then the cosets of $N$ are the cells of a partition of $G$ in which we can define the coset equivalences described above with $xN=yN\Rightarrow x\sim y$. Proof using the right coset follows by symmetry.

Last edited: Sep 15, 2015
7. Sep 15, 2015

### micromass

Staff Emeritus
Seems right, so allow me now to give some background information concerning this problem.
The classical treatment of groups defines quotients by considering a normal subgroup and "making that vanish". In the same way, in the theory of rings, we consider ideals and "making it vanish". So in both cases, we consider a special subset and "make it vanish".

But when considering other algebraic structures, we cannot seem to do that. For example, semigroups and lattices don't have easily detectable subsets which we can make vanish. The answer lies with congruence relation. A congruence relation is an equivalence relation respecting the operation. The equivalence relation in your OP is a congruence. Now it turns out that when taking quotients, the congruence relation is the fundamental object. Congruence relations on semigroups and lattices are easily defined and lead to a very good notion of quotient structures (with their own version of the isomorphism theorems!). As you've shown now, the congruence relation approach to quotients and the normal subgroup approach to quotients are equivalent in groups. In rings, we have the same situation: congruence relations and ideals are equivalent. This is where normal subgroups and ideals come from.

This provides a link also with other quotients like in topology, where we quotient out a relation, and not necessarily a subset.