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

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SUMMARY

The discussion focuses on the structure of the finite quotient group G/H, where G is the cyclic group Z6 and H is the normal subgroup (0,3). It is established that the cosets of H in G are H, H+1, and H+2, leading to three distinct cosets despite having multiple representations. The equivalence of cosets is illustrated by showing that H+3 is identical to H, confirming the properties of quotient groups. The discussion concludes that Z6 can be expressed as a union of distinct cosets of H, reinforcing the cyclic nature of the group.

PREREQUISITES
  • Understanding of group theory concepts, specifically quotient groups.
  • Familiarity with cyclic groups, particularly Z6.
  • Knowledge of normal subgroups and their properties.
  • Basic comprehension of set notation and equivalence relations.
NEXT STEPS
  • Study the properties of normal subgroups in group theory.
  • Learn about isomorphisms and their applications in group theory.
  • Explore the concept of cosets in greater detail, including left and right cosets.
  • Investigate the geometric interpretation of groups and cosets using complex numbers.
USEFUL FOR

Mathematicians, students of abstract algebra, and anyone interested in the structure and properties of groups and quotient groups.

onie mti
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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?
 
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onie mti said:
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.
 
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).
 

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