@KannappanSampath Ok. I'm here.

Exercise 1, page 45: Prove that $Z_n^\ast$ is a group.

The definition of $Z_n^\ast$ is given on page 2 in examples (1.2) (5): $Z_n^\ast := \{ m \mid 1 \leq m < n \text{ and } m \text{ relatively prime to } n \}$

The operation is multiplication modulo $n$.

Now we'd like to verify that multiplication is associative, that we have a neutral element and that every element has an inverse.

Our elements are $\{\bar 1, \bar 2, \cdots, \overline {n-1}\}$.

So, we claim that $\bar 1$ is the identity.

Yes.

$\bar{1}$ is the neutral element of the group since $\bar{1} \cdot \bar{a} = \overline{1 \cdot a} = \bar{a} $.

What do you think of this?

Yes, right!

Really? It feels shaky for me. It's been ages since the last time I looked at groups.

It is right, because, we just have $1$ times an element less than $n-1$. So,...

Yes.

Now we want inverses and associativity.

For instance in $\Bbb Z_5^\ast$, $\bar 2 \cdot \bar 3=\bar 6=\bar 1$. Here, the problem is with representatives.

Right.

Now inverses next: let $\bar{a} \in Z_n^\ast$.

We want $b \in Z_n^\ast$ such that $\overline{ab} = \bar{1}$.

Since the elements are co-prime to $n$, we can make use of Euclid's algorithm in disguise!

So, for $a$ co-prime to $n$, there exists, $b$ and $c$ in $\Bbb Z$ such that, $ab+nc=1$. Right?

Now, reading that equation $\mod n$, we see, $\overline{ab}=\overline 1$.

Yes, looks good.

Next associativity. Let $\bar{a}, \bar{b}, \bar{c} \in Z_n^\ast$.

For associativity, $\bar a \cdot (\bar b \cdot \bar c)=\bar a \cdot \overline{bc}=\overline{a \cdot bc}$

$= \overline{abc} = \overline{ab} \cdot \bar{c} = (\bar{a} \cdot \bar{b}) \cdot \bar{c}$.

Yes, cute!

: )

So, we are through with this exercise!

Yes.

Exercise 2, page 45: Prove $P(X)$ with symmetric difference is a group.

(where $P(X)$ is the power set of an arbitrary set $X$)

$P(X)$ here is used to denote the power set and for $A,B \in P(X)$ we have the symmetric difference $A \Delta B := (A \setminus B) \cup (B \setminus A)$.

Yes. So, again the set of axioms to verify: Associativity, identity, inverse.

Note that $\varnothing $ is the neutral element since $A \Delta \varnothing = (A \setminus \varnothing ) \cup (\varnothing \setminus A) = A \cup \varnothing = A$.

What do you think?

Sure, right.

And, every element is its own inverse.

True.

Now for associativity we need to show $A \Delta (B \Delta C) = (A \Delta B) \Delta C$.

Because, $A \Delta A=(A \setminus A )\cup( A \setminus A)=\varnothing$, which is the identity.

Yep.

Associativity is a bit nasty to verify. :/

$$ A \Delta ((B \setminus C) \cup (C \setminus B)) = (A \setminus ((B \setminus C) \cup (C \setminus B))) \cup (((B \setminus C) \cup (C \setminus B)) \setminus A)$$

But how go we simplify this? It scares me off!

I'm working on it : )

$$ (A \setminus ((B \setminus C) \cup (C \setminus B))) \cup (((B \setminus C) \cup (C \setminus B)) \setminus A) = (A \cap ((B \setminus C) \cup (C \setminus B))^c) \cup (((B \setminus C) \cup (C \setminus B)) \cap A^c)$$

$$ = (A \cap ((B \setminus C)^c \cap (C \setminus B)^c)) \cup (((B \setminus C) \cup (C \setminus B)) \cap A^c) $$

Now let's see what we want on the RHS:

$$ (A \Delta B) \Delta C = (((A \setminus B) \cup (B \setminus A)) \setminus C) \cup (C \setminus ((A \setminus B) \cup (B \setminus A))) $$

$$ = (((A \setminus B) \cup (B \setminus A)) \cap C^c) \cup (C \cap ((A \setminus B) \cup (B \setminus A))^c) $$

$$ = (((A \setminus B) \cup (B \setminus A)) \cap C^c) \cup (C \cap ((A \setminus B)^c \cap (B \setminus A)^c))$$

$$ = (((A \cap B^c) \cup (B \cap A^c)) \cap C^c) \cup (C \cap ((A \cap B^c)^c \cap (B \cap A^c)^c))$$

I think we must simplify those $A\setminus B$ as well. What do you think?

Yes. : S

What a pain.

This is boring. Shall we skip it?

And, still how do you deal with what comes? I am not seeing how we'll prove.

Ok. Let me do it on paper and then post it here.

@MattN Thank you, Matt. I'll try here as well!

@MattN I think I am now seeing it. I'll do this and post a .pdf later tommorrow. I think we can skip it. Or, have you finished working it?

(Sorry to disappoint, if you have written it out neatly and whatnot!)

@KannappanSampath Not yet. But it's not difficult, it's just very boring : )

That's ok.

So, we shall skip it now?

Yes.

Exercise 3, page 45: Write the Cayley diagram for the group $S_3$.

That's the Cayley Graph of $S_3$ right?

Apparently. : )

I wonder what the green lines represent.

Here is a more informative link about $S_3$ in particular about the Cayley graph and Cayley table.

And, that does not match with the Cayley Graph of $S_3$ here in the link

Here's a better picture:

Taken from here.

Now, the graph is easier to understand because, we have the generating set. : )

I need start cooking dinner.

Now, what do you say we skip this and the next as well?

Sure.

@MattN Sure. I'll see you later. By this time, I'll TeX the exercises we have discussed.

We could skip all the group stuff and directly start with rings.

: )

Ah, precisely, I thought too. But, you said you wanted to do group theory. So,.. :-)

I thought the exercises would be different : )

@KannappanSampath Hello. Now I'm back but not in the mood to do any more algebra. : (

@MattN : ( Fine. I am reading anyways, as you might have had a course. So, you'll catch up faster I know.

@KannappanSampath Sorry. I'm a lazy person : (

@MattN Never mind. Going to do Analysis now?

@KannappanSampath I was planning to do nothing at all.

@MattN Oh, since it is a weekend, I assume it is movie time now. I am going to watch a movie after an hour or so!

@KannappanSampath Let me know what you thought of Incredibles, once you've watched it : )

@KannappanSampath Actually, I'm so tired that I'm not even in the mood to watch a film.

@MattN Sure! When next, then, will you be around?

@KannappanSampath Tomorrow morning.

That's in about 12 hours from now.

@MattN I think you must get some sleep :p

@MattN Oh, I'll be here then, I guess.

@Matt Take rest. Hope to see you fresh and in a good mood for Algebra. : )

@KannappanSampath Will try : )