Group Theory - 2022-05-22

Jack Rod

09:17

actually I am confuse in some groups aH=Ha

and In some they are not equal

Martin Sleziak

← 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.

Jack Rod

one basic doubt

like I have an example set h={(1),(1,2,3)}

Martin Sleziak

ok

Jack Rod

let a=(2,3)

Martin Sleziak

do you mean set of permutations? I am not sure what your notation (1,2,3) means.

Jack Rod

09:24

yes probably

cyclic permutations

Martin Sleziak

Is (1,2,3) a permutation? Is this your notation for 3-cycle $1\mapsto2\mapsto3\mapsto1$?

Jack Rod

yes

Martin Sleziak

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).

Jack Rod

what is $S_3$

Martin Sleziak

$S_3$= 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 $aH=Ha=\{(12),(13),(23)\}=G\setminus H$.

Jack Rod

09:29

a set H can be sub group of G only if it is closed ,and bijected?

Martin Sleziak

I do not know what "bijected" means.

Jack Rod

one one and onto

I was writing bijection

Martin Sleziak

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.)

Jack Rod

we can explain same thing by quotient group?

Martin Sleziak

I don't understand the question.

Jack Rod

09:35

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.)

Martin Sleziak

If H is a normal subgroup, then it is possible to form the quotient group G/H.

Jack Rod

where is H is the same u have written

a=(2,3)

Martin Sleziak

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 $Ha$.

Jack Rod

ok got it

last one theorem of factor/quotient

Martin Sleziak

Although in this case, we know that without computation, too - just from the number of elements: Given $G$ group and $H \le G$ such that $|G:H|=2$, how does $x^2 \in H$ for every $x \in G$?, if $H \leq G$ has index 2, then $a^2\in H$ for every $a\in G$, Subgroup of Index 2 is Normal

Jack Rod

09:39

HaHb=Hab

a,b belong to G

Martin Sleziak

Ok, what is the question?

Jack Rod

where H is subgroup of G

Martin Sleziak

Is H a normal subgroup? And what is your question?

Jack Rod

yes H is normal subgroup

Martin Sleziak

ok.

And what is the question you want to ask about this?

Jack Rod

09:41

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

Martin Sleziak

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.)

Jack Rod

do u have nay other

please tell

I very confuse in quotient group

Martin Sleziak

What you mean by "please tell"? Other possibilities to define the quotient group?

Jack Rod

yes

bookish definition is like a bouncer

Martin Sleziak

One could define a quotient group using congruence relations. Or one could use different meaning for $Ha\cdot Hb$ - namely as multiplication of subsets.

Jack Rod

09:47

I am reading group theory on my own for first time

Martin Sleziak

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.

Jack Rod

joseph.A gallian

Martin Sleziak

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?

Jack Rod

yes

Martin Sleziak

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.

Jack Rod

09:56

ok

Martin Sleziak

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.

Jack Rod

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

Martin Sleziak

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.

Jack Rod

find the quotient group g/h

when g=

(z__12,+12) and h=={(0,3,6,9)}

and also prepare compostion table

@MartinSleziak

Martin Sleziak

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.)

Martin Sleziak

11:14

@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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$Mathematics$ Group Theory