Some very short clarification on quotient groups

Question

This is something I'm slightly confused with: I have a group, $S_3$, say, and its normal subgroups $N$ are $\{e\}, A_3$ and $S_3$. Then its quotient groups are thus $S_3/N$ for each $N$. But what specifically are these groups? I'm quite sure $S_3/\{e\} = \{e, (12), (13), (23), (123), (132)\} = S_3$, but is this right? If so, what is $S_3/A_3$?

Answer 1

Since $S_3/A_3$ is rather small, let's just zoom out and focus a bit more on the general situation of $N \lhd G$, and what exactly happens in $G/N$, algebraically.

When $N \lhd G$, the elements of $G/N$ are just cosets of $N$, so we have

$$G/N = \{gN:g \in G\}.$$

The operation in $G/N$ is the operation of $G$, but on the level of cosets: Given $g, h \in G$, we have $gN, hN \in G/H$, and coset multiplication is (well-)defined as

$$(gN)(hN) = (gh)N = ghN.$$

We literally multiply cosets of $N$ by multiplying their coset representatives, which gives us a coset representative of their product.

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Examples

When $G = S_3$ and $N = A_3$, our only cosets are indeed $A_3$ and $(12)A_3$. When we multiply $A_3$ by anything, it acts as the identity of $S_3/A_3$,

$$(gA_3)(eA_3) = (ge)A_3 = gA_3 = (eg)A_3 = (eA_3)(gA_3),$$

while $(12)A_3$ is an element of order $2$, since

$$\big((12)A_3\big)^2 = \big((12)A_3\big)\big((12)A_3\big) = (12^2)A_3 = eA_3 = A_3$$

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In the case $G = A_4$, we have a normal subgroup $N = \{e,(12)(34),(13)(24),(14)(23)\} \lhd A_4,$ so that $A_4 / N = \{N, (123)N, (123)^2N\}$, and $A_4/N$ behaves just like a cyclic group with three elements.

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Considering $Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$ is the quaternion group, then $N = \{\pm 1\} \lhd Q_8$.

The quotient group $Q_8 / N = \{N, iN, jN, kN\}$ is isomorphic to the Klein $4$-group, where, for example, $(iN)(jN) = (ij)N = kN$.

At any point you could consider any of these coset $gN$ as a subset of the original group, i.e., $kN = \{k, -k\} \subseteq Q_8$, but it's generally more enlightening and less confusing (for me at least) to just think of it as some 'translate' (coset) of the normal subgroup $N$.

Answer 2

As it is known, if $t_1=1,\ldots,t_m$ - transversal of $N\unlhd G$, then $G/N=\{N,t_2N,\ldots,t_mN\}$ and all classes $N,t_2N,\ldots,t_mN$ are distinct. For example, if $N=\{e\}$, then $G$ - transversal of $N$ and
$$ S_3/\{e\}=\{\{e\},\{(1,2)\},\{(1,3)\},\{(2,3)\},\{(1,2,3)\},\{(1,3,2)\}\}. $$ If $N=A_3$, then $1,(1,2)$ - transversal of $N$, so $S_3/A_3=\{A_3,(1,2)A_3\}$.