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9:17 AM
actually I am confuse in some groups aH=Ha
and In some they are not equal
 
2 messages moved from Martin Sleziak's room
The condition that aH=Ha holds for every a∈G holds for normal subgroups.
A subgroup H is normal if and only if the left cosets are the same as the right cosets.
 
one basic doubt
like I have an example set h={(1),(1,2,3)}
 
let a=(2,3)
 
do you mean set of permutations? I am not sure what your notation (1,2,3) means.
 
9:24 AM
yes probably
cyclic permutations
 
Is (1,2,3) a permutation? Is this your notation for 3-cycle $1\mapsto2\mapsto3\mapsto1$?
 
yes
 
I should point out that your h isn't a subgroup of $S_3$.
Typically we look at cosets $aH$, $Ha$ in cases when $H$ is a subgroup (of some group).
 
what is $S_3$
 
$S_3$= group of all permutations of 1,2,3 (and the group operations is composition)
If you take the smallest subgroup containing the 3-cycle (123), you get $H=\{id,(123),(132)\}$. (The alternating group.)
This subgroup is normal in $S_3$ simply because $[G:H]=2$.
If $a\in H$, then the cosets are $aH=Ha=H$.
And if $a\in G\setminus H$, then the cosets are $aH=Ha=\{(12),(13),(23)\}=G\setminus H$.
 
9:29 AM
a set H can be sub group of G only if it is closed ,and bijected?
 
I do not know what "bijected" means.
 
one one and onto
I was writing bijection
 
You can say about a function that it is (or isn't) a bijection.
I do not know what you mean when you say that a subset is a bijection.
If $G$ is a group and $H\subseteq G$ is a non-empty subset, then it is a subgroup
if and only if it is closed both under the operation and under the inverses.
$$(\forall a,b\in H) ab\in H, a^{-1}\in H$$
Or you can put these conditions into one: $$(\forall a,b\in H)ab^{-1}\in H.$$
However, if $H$ is finite, it is enough to verify that it is closed under the group operation. (It follows automatically that it is closed under inverses, too.)
 
we can explain same thing by quotient group?
 
I don't understand the question.
 
9:35 AM
aH=Ha
9 mins ago, by Martin Sleziak
If you take the smallest subgroup containing the 3-cycle (123), you get $H=\{id,(123),(132)\}$. (The alternating group.)
 
If H is a normal subgroup, then it is possible to form the quotient group G/H.
 
where is H is the same u have written
a=(2,3)
 
I am still in dark what you're actually asking.
For $G=S_3$, $H$ as above and $a=(23)$, you'll get $aH=Ha$.
One can verify that by computing the permutations belonging to $aH$ and $Ha$.
 
ok got it
last one theorem of factor/quotient
 
9:39 AM
HaHb=Hab
a,b belong to G
 
Ok, what is the question?
 
where H is subgroup of G
 
Is H a normal subgroup? And what is your question?
 
yes H is normal subgroup
 
ok.
And what is the question you want to ask about this?
 
9:41 AM
question set of all cosets of a normal subgroup H in
a group G
is a group with respect to multiplication of cosets
defined HaHb=Hab
where a,b belong to G
 
What you wrote is one possibility how one can define the quotient group. (If we define it in this way, we'll need to show that the operation is well-defined.)
 
do u have nay other
please tell
I very confuse in quotient group
 
What you mean by "please tell"? Other possibilities to define the quotient group?
 
yes
bookish definition is like a bouncer
 
One could define a quotient group using congruence relations. Or one could use different meaning for $Ha\cdot Hb$ - namely as multiplication of subsets.
 
9:47 AM
I am reading group theory on my own for first time
 
But I would recommend to stick with the definition you've been using. I am afraid it will only get more confusing if several various definitions are thrown at you at the same time.
@JackRod And what text are you reading?
Moreover, I am afraid that explaining congruence relations would take a long time.
 
joseph.A gallian
 
Of course, if you prefer to check the congruences, you can find various explanations online.
For example, this is on YouTube: (Steven Roman Mathematics) Normal Subgroups, Quotient groups and Congruence Relations
Gallian: Contemporary Abstract Algebra?
 
yes
 
This is the first occurrence of the quotient group that I see in that book.
And it is followed by several examples where the quotient group is finite.
 
9:56 AM
ok
 
Actually, I think Z/nZ is a reasonable example to get acquainted with quotient groups.
Moreover, it is something we commonly use - modular arithmetic, days in a week, hours in a day.
 
what I got from def is
if a group like whose elements are like n sided polygons
each are evenly spapced
@MartinSleziak I have an eg
if u can work out that it will be benifical
 
Ok, and what is your example?
I might have to leave soon. But even if somebody else looks here, it might be good idea to state clearly what is the example you're interested in.
I.e., what is the group G and what is the normal subgroup H.
 
find the quotient group g/h
when g=
(z__12,+12) and h=={(0,3,6,9)}
and also prepare compostion table
@MartinSleziak
 
Sorry, I'll have to leave. I might have a look at this room later. (But maybe somebody might notice that there was a new activity in chat and respond here.)
 
 
1 hour later…
11:14 AM
@JackRod Sorry, I had to leave - some stuff IRL.
Any case, this should not really be that difficult - in fact, it is a bit similar to the example with Z/4Z which is one of the examples in Gallian's book. (Although this is closer to 3Z.)
First thing first.
H={0,3,6,9} is a subgroup of Z_{12}. (BTW you have a included redundant brackets there by mistake.)
The group Z_{12} is commutative - so every subgroup is normal.
We get three cosets \begin{align*}
0+H&=\{0,3,6,9\}\\
1+H&=\{1,4,7,10\}\\
2+H&=\{2,5,8,11\}\end{align*}
And then we can form the group table which looks rather similar to $\mathbb Z_3$.
\begin{array}{c|ccc}
+ & 0+H & 1+H & 2+H \\\hline
0+H & 0+H & 1+H & 2+H \\
1+H & 1+H & 2+H & 0+H \\
2+H & 2+H & 0+H & 1+H
\end{array}
So in this case we have $G/H\cong (\mathbb Z_3,+_3)$.
As a side note, once we know that $|G/H|=3$, this already means that it is isomorphic to $\mathbb Z_3$.
In the part about Lagrange's theorem, Gallian mentions that any group of prime order is cyclic.
And a cyclic group of order $n$ is isomorphic to $\mathbb Z_n$.
Of course, the fact that 3 is a prime number and thus we know this is Z3 is only tangential here - you have asked about this mainly as an example of a quotient group.
For the benefit of others, I'll include a link to another room, where you have a discussion about quotient groups: chat.stackexchange.com/transcript/111475/2022/5/22
 

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