MHB Understanding Finite Quotient Groups: G/H with G=Z6 and H=(0,3)

[onie mti]

G is a group and H is a normal subgroup of G.
where G=Z₆ and H=(0,3)
i was told to list the elements of G/H
I had:
H= H+0={0,3}
H+1={14}
H+2={2,5}
now they are saying H+3 is the same as H+0, how so?

 

[I like Serena]

G is a group and H is a normal subgroup of G.
where G=Z₆ and H=(0,3)
i was told to list the elements of G/H
I had:
H= H+0={0,3}
H+1={14}
H+2={2,5}
now they are saying H+3 is the same as H+0, how so?

Yes. That is the idea of a quotient group.
The difference between 0 and 3 is "divided" out.
They are considered equivalent.

Alternatively, since G has 6 elements and we divide by a H with 2 elements, that means that G/H has 6/2=3 elements.

 

[Deveno]

Let's just do the math:

What is $H+3$?

Well, $H+3 = \{g \in G: g = h+3, h \in H\}$

So we add 3 to every element of $H = \{0,3\}$.

Thus $H+3 = \{0+3,3+3\}$.

However, in $\Bbb Z_6$, we have $3+3 = 0$, because 6 is a multiple of 6.

Thus, $H+3 = \{3,0\} = \{0,3\} = H$ (since in sets, the order in which we list the elements doesn't matter).

Similarly, $H+4 = \{0+4,3+4\} = \{4,1\} = \{1,4\} = \{0+1,3+1\} = H+1$.

I think you can convince yourself that $H+5 = H+2$.

Thus, out of the 6 possible "conceivable" cosets we have, it turns out that:

$H = H+3$
$H+1 = H+4$
$H+2 = H+5$

so there are only 3 DISTINCT cosets, although each one of the 3 has "two different names".

This is something to be wary of with cosets: a coset $Hg$ can have many different $g$ that identify the same coset. So even if $g \neq g'$, it still might be that $Hg = Hg'$.

IN GENERAL, if $Ha = \{a,h_1a,\dots,h_na\}$, we have $n+1$ different "names" for the same coset:

$Ha = Ha_k$ where $a_k = h_ka$.

Another interesting fact (which turns out always to be true):

$\Bbb Z_6 = \{0,1,2,3,4,5\} = \{0,3\} \cup \{1,4\} \cup \{2,5\} = H \cup (H+1) \cup (H+2)$,

any group $G$ is a union of the distinct cosets of any subgroup $H$.

In plane geometry, a coset of a line $L$ through the origin, is a coset $(x_0,y_0) + L$, which is a line parallel to $L$ that goes through the point $(x_0,y_0)$. This new line (the coset) is often called "the translation of $L$ by $(x_0,y_0)$" which represents "how much we shift $L$".

By analogy, cosets $Hg$ are often (especially in the ABELIAN case) called (in this case right) translations of $H$ by $g$. In other words, we can think of $H+k$ as "shifting $H$ by $k$ units", and doing this, it is clear for $\Bbb Z_6$ and $H = \{0,3\}$ we only have 3 possible shifts:

The null shift (shift by 0)
Shift by 1 (or 4, or 7, or...)
Shift by 2 (or 5, or 8, or...)

before the cycle repeats itself. That's why $\Bbb Z_6$ is called "cyclic", and is best visualized as the 6 vertices of a regular polygon (this can be done explicitly if one takes:

$[k] \iff \cos(2\pi k/6) + i\sin(2\pi k/6)$

in the complex numbers, and uses complex multiplication, instead of addition mod 6. The above is an example of an ISOMORPHISM, something you will soon be reading very much about).