Is my solution for this exercise correct?

[evinda]

Hello! (Smirk)

I am given this exercise:

$$\text{Let } \phi: \mathbb{Z} \to m \mathbb{Z}, \text{ that is defined like that: } \phi(a)=ma$$
a) $\phi(a)=ma \text{ is a group homomorphism }$
b) $\phi(a)=ma \text{ is a ring homomorphism }$
c) $\text{the group } \mathbb{Z}_2 \times \mathbb{Z}_4 \text{ is cyclic}$
d)$\text{the group } \mathbb{Z}_4 \times \mathbb{Z}_3 \text{ is cyclic}$
e)$\text{ the groups } D_4 \text{ and } \mathbb{Z}_8 \text{ are isomorphic}$
f) $\text{ the groups } \mathbb{Z}_4 \text{ and } U_4 \text{ are isomorphic}$

For each sentence $p \in \{a,b,c,d,e,f \}$,let $t(p)=1$ if $p$ is true, $t(p)=-1$ if $p$ is false.

Calculate $5t(a)+13t(b)-17t(c)+5t(d)+7t(e)+9t(f)$

I thought that it is equal to : $5-13+17+5-7+9=20+9-13=29-13=16$

(a->true,b->false,c->false,d->true,e->false,f->true)

Could you tell me if it is right? (Sweating)(Sweating)

 

[Deveno]

Yes, I could tell you.

But here is what I think would be better: I would like for you to tell me if YOU are right, and why.

Let me explain. You seem unsure of whether or not you are doing this correctly. That indicates to me that you don't know. But if something is in fact true, there should be no question. True is true. If you don't know whether or not you understand what you have learned, you haven't actually learned it, yet. This is something that is actually USEFUL: if you want to understand what you have learned more deeply, try explaining it to someone.

And if you HAVE learned it, you ought to be certain in such knowledge. A young child may be uncertain if 2 and 2 is always 4, but an accountant should not be so uncertain, or she will not be an accountant for very long.

If $U_4$ is meant to be the group of units modulo 4, I think you should take a closer look at (f).

Math is not about "getting the right answer". I suspect you are a person who seeks to please, that you are in general well-mannered and polite, so you may be accustomed to asking people: "Is this OK?"

But math is about ideas, and using those ideas as TOOLS to solve problems we might not know the answers to in advance. So using these tools needs to come as naturally to you, as a carpenter swinging a hammer. For if a carpenter does not swing true, it's a little hard on his thumbs.

I am not your teacher, so I do not know how you are being evaluated, but if I was your teacher, I would give "more marks" for wrong answers with the right logic behind them, then for the right answers with no explanation given.

 

[I like Serena]

Hi! (Smile)

What is $U_4$? (Wondering)

 

[evinda]

Hi! (Smile)

What is $U_4$? (Wondering)

$U_4=<\zeta: 1, \zeta, \zeta^2, \zeta^3>$ .. (Thinking)

 

[I like Serena]

$U_4=<\zeta: 1, \zeta, \zeta^2, \zeta^3>$ .. (Thinking)

I'm going to go ahead and say that I believe it is all correct. (Sun)

Btw, does this $U_4$ have a name? (Wondering)
I've never seen it before.
I do know $C_4$, which represents the cyclic group of 4 elements, which is apparently exactly what this $U_4$ is.
Anyway, the only group with a $U$ in it that I know is $U(n)$, which is the group of unitary nxn matrices.

 

[Deveno]

Perhaps what is meant is the group of 4th roots of unity:

$\{i,i^2 = -1,i^3= -i,i^4 = 1\}$

 

[evinda]

I'm going to go ahead and say that I believe it is all correct. (Sun)

Great! Thank you very much! (Tongueout)

Btw, does this $U_4$ have a name? (Wondering)
I've never seen it before.
I do know $C_4$, which represents the cyclic group of 4 elements, which is apparently exactly what this $U_4$ is.
Anyway, the only group with a $U$ in it that I know is $U(n)$, which is the group of unitary nxn matrices.

Perhaps what is meant is the group of 4th roots of unity:

$\{i,i^2 = -1,i^3= -i,i^4 = 1\}$

Yes,that's what I mean with $U_4$.. (Nod)(Nod)

 

[Deveno]

In that case, it appears that your answer is correct.

a) Is obvious because of the distributive law.

b) $\phi(ab) = m(ab) \neq m^2(ab) = (ma)(mb) = \phi(a)\phi(b)$, unless $m = 0$ or $1$.

c) This group has no element of order 8 (it's EXPONENT, the smallest power which makes every element the identity, is 4)

d) gcd(3,4) = 1 (Chinese Remainder Theorem)

e) $D_4$ is not abelian

f) any two cyclic groups of the same cardinality are isomorphic, for example $k \cdot 1 \mapsto (i^3)^k$ is an isomorphism here.

 

[evinda]

In that case, it appears that your answer is correct.

a) Is obvious because of the distributive law.

b) $\phi(ab) = m(ab) \neq m^2(ab) = (ma)(mb) = \phi(a)\phi(b)$, unless $m = 0$ or $1$.

c) This group has no element of order 8 (it's EXPONENT, the smallest power which makes every element the identity, is 4)

d) gcd(3,4) = 1 (Chinese Remainder Theorem)

e) $D_4$ is not abelian

f) any two cyclic groups of the same cardinality are isomorphic, for example $k \cdot 1 \mapsto (i^3)^k$ is an isomorphism here.

Nice,thanks a lot! :)