Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

EDIT: I'm only allowed to ask 6 questions and as my next question is similar to this, I thought I'd post it in this thread. The first question is the new one. The one under the underlined bit is the old question:

I tried it with looking at the previous posts. Now I'm trying to write it up and explain without looking at them.

I have

$$H = C_7 = \langle a \rangle \hspace{2cm} Q = C_3 = \langle b \rangle$$

To find my SDP's, I want to define a homomorphism, $\theta: Q \rightarrow \mathrm{Aut}(H)$. This is given by $\theta: C_3 \rightarrow C_6$. We know that non-trivial SDP's exist as $3$ divides $6$ and so we can see that we get $3$ SDP's altogether: $1$ trivial and $2$ non trivial.

Let $\mu$ generate $\mathrm{Aut}(C_7)$. I know want to define some $\mu$ of some order $r$ that sends $b \rightarrow b^{d^r} = b$. I.e I want to find some $d^r \equiv 1 \mod 7$. To find my $d$, I first split $6 = 2 \times 3$. Here we see that gcd$(2,7) = 1$ and so this $r$ exists for some $d$. We then get

$$2^1 \equiv -5 \mod 7$$ $$2^2 \equiv -3 \mod 7$$ $$2^3 \equiv 1 \mod 7$$

Therefore, I get that $\mu: b \mapsto b^3$.

Now, I have this, I can now define two multiplication orders (for my two non trivial SDP's) and construct the SDP's. As $\mathrm{Aut}(C_7)$ is cyclic, we get that $C_6 = \langle \mu \rangle$.From here, out two elements of order $r = 3$ are $\mu^2$ and $\mu^4$. Using the multiplication order

$$ ba = a\theta(a^{-1})(b)$$

where $\theta(a^{-1})(b) = \mu^{2,4}$, we get our first SDP to be when we map $\mu^2$.

$$\mu: b \mapsto b^3 \implies \mu^2: b \mapsto b^9$$

$9 \equiv 2 \mod 7$ and so we get $\mu^2: b \mapsto b^2$ and so our first multiplictive order is defined as $ba = ab^2$.

For the second one, we know have $\mu^4 : b \mapsto b^{3^4}$. $3^4 = 81 \equiv 4 \mod 7$ and so we get that $\mu^4 : b \mapsto b^4$. So our second multiplicative order is defined as $ba = ab^4$.

From here, we conclude that we have 3 SDP's. One is the direct product so

$$G \cong H \times Q \cong C_6 \times C_3$$

We then have two more groups whose elements follow the multiplicative orders $ba = ab^2$ and $ba = ab^4$.

Is this correct?


Hopefully this will be the last SDP question I post because I get how to construct them. I don't have an answer to this as well because I've adjusted the question I was asked (how many SDPs are there) to find the SDP. This is what I did:

$$H = C_{42} \hspace{2cm} Q = C_3$$

Let $H = \langle a \rangle, Q = \langle b \rangle, \mathrm{Aut}(H) = \langle \mu \rangle$.

An SDP is a homomorphism, $\theta : Q \rightarrow \mathrm{Aut}(H)$. So my first step is to calculate $\mathrm{Aut}(H)$. $42 = 2 \times 3 \times 7$. So this gives us $\mathrm{Aut}(H) \cong C_2 \times C_6 \cong C_{12}$. As $3$ divides $12$, we know there is a nontrivial SDP.

We know that $\mu$ generates $\mathrm{Aut}(H)$ and so we want to find an element in $\mathrm{Aut}(H)$ which has order $12$. We can work this out and say $\mu^3$ is one of these elements.

We want to define some $\mu : b \mapsto b^k$. We can say that therefore $k = 2$ as two $2$ is the lowest integer whose multiplicative integer is $\equiv 2 \mod 12$. So now we have definied $\mu: b \mapsto b^2$.

We want to work out the multiplicative order

$$ba = a\theta(a^{-1})(b)$$

Where $\theta(a^{-1})(b)$ generates $\mu$. Seeing as we definied $\mu : b \mapsto b^2$, we therefore get $\mu^3: b \mapsto b^{2^3} = b^8$ and so we get our SDP as a group with multiplicative order

$$ba = ab^8.$$

Right now thats what i've got. I'm still working through the next bit to write it the $(a^i, b^j)$ form but my lecturer hasn't been writing it like that so I think the multiplicative order answer would be ok in the exam.

share|improve this question
2  
Don't rush it: $\,C_2\times C_6\ncong C_{12}\, $! –  DonAntonio Jan 12 '13 at 13:24
    
@DonAntonio Oh ok, so then I want elements of order $3$ in $C_2 \times C_6$. There are none in $C_2$ but in $C_6$ I have some and so I can define $\mu$ as $\mu^2$. Then blah blah blah and I get my multiplicative order as $ba = ba^4$? –  Kaish Jan 12 '13 at 14:01
    
That should be $ba = ab^4$ –  Kaish Jan 12 '13 at 14:06
    
Indeed, @Kaish...or, as adviced before, use ordered pairs. :) –  DonAntonio Jan 12 '13 at 14:23
    
@DonAntonio Lol thanks. Yeah, I will defo look at them. I might not ask as many question on them because I think people are getting annoyed with my constant SDP questions, but if I can put them in that form, atleast then I will be able to see the group easier for myself –  Kaish Jan 12 '13 at 14:30
show 6 more comments

1 Answer

up vote 2 down vote accepted

First question:

To find elements of $\text{Aut}(C_{42})$ look for $x\mapsto x^d$ for $d$ relatively prime to $42$. If you compose this map with itself, you get $x\mapsto (x^d)^d=x^{d^2}$, if you do it again you get $x\mapsto x^{d^3}$ and so on. If you're looking for an automorphism of order $r$, what you want is $x\mapsto x^{d^r}=x$, which means that $d^r\equiv 1 \pmod{42}$.

Let's hold up one sec: why does $d^r\equiv 1 \pmod{42}$?

Every element has order dividing $42$ in $C_{42}$. In particular, $1$ has order $42$, so if $x^{d^r}$ is supposed to be equal to $x$ for every $x\in C_{42}$, including $1$, we need for $d^r$ to be some multiple of $42$ plus one.

So, you want to find an element of order $3$ in $\text{Aut}(C_{42})$. You know that one exists by Cauchy's theorem, since $|\text{Aut}(C_{42})|=12$. Let's start by looking at maps $x\mapsto x^d$ for different $d$ and see what orders we can get.

$42=2\times 3 \times 7$ so the smallest number for which $\text{gcd}(d,42)=1$ is $d=5$. Let's find the order of $x\mapsto x^5$ in $\text{Aut}(C_{42})$: $$ \begin{eqnarray*} 5&\equiv&5\pmod{42}\\ 5^2&\equiv&25\pmod{42}\\ 5^3&\equiv&41\pmod{42}\\ \end{eqnarray*} $$ Whoa whoa whoa - so $5^3\equiv-1\pmod{42}$? Recall my comment on DonAntonio's answer in your other thread on multiplicative order - this means that $5$ has multiplicative order $6$ modulo $42$.

Thus $x\mapsto x^5$ has order $6$ in $\text{Aut}(C_{42})$. But so what? We were looking for an automorphism of order $3$. Well, all you have to do to turn an element of order $6$ into an element of order $3$ is square it. So $x\mapsto x^{25}$ has order $3$ in $\text{Aut}(C_{42})$.

Now that you explicitly have the element of order $3$ in $\text{Aut}(C_{42})$ I think you can finish the problem.


Second question:

You are mostly correct, but there are a couple things:

When you're doing the $d^r\equiv 1 \pmod{n}$ thing, you are finding the multiplicative order of $d$ mod $n$. Equivalently, $x\mapsto x^d$ is an automorphism of order $r$ in $\text{Aut}(C_n)$ (assuming $\text{gcd}(d,n)=1$). So, $\mu$ is a generator of $\text{Aut}(C_p)$ if and only if $\mu:x\mapsto x^d$ where $d$ has order $p-1$ mod $p$.

Here what you want to do is find some automorphism of order $3$ to map $a$ to. You correctly found that the multiplicative order of $2$ is $3$ mod $7$, so you have that $x\mapsto x^2$ is an automorphism of order $3$ in $\text{Aut}(C_7)$.

From there you determine that the generator of $\text{Aut}(C_7)$ is $\mu:x\mapsto x^3$, which is true but I'm not sure how you got it from the preceding computation. If you noticed that $3^2\equiv 2 \pmod{7}$ then your reasoning was correct, but if you concluded that $d^r\equiv 1 \pmod{7}$ implies that $x\mapsto x^r$ generates $\text{Aut}(C_7)$, that is not true.

With that said, the rest of your computations are correct; indeed $\langle a,b|a^3,b^7,b^a=b^2\rangle$ and $\langle a,b|a^3,b^7,b^a=b^4\rangle$ are the correct semidirect products. As it turns out, however, these are isomorphic, so actually you have only two distinct SDPs (including the trivial semidirect product $C_7\times C_2$).

share|improve this answer
    
From there, I can still use my $\mu^2$ and then I can get my multiplication order to be $ba = ab^{25^2}$? –  Kaish Jan 12 '13 at 17:14
1  
You're good with just $ba=ab^{25}$. In this case the $\mu$ is just $x\mapsto x^5$ so $\mu^2$ is $x\mapsto x^{25}$. (That is what I was saying in the $2^\text{nd}$ to last paragraph.) –  Alexander Gruber Jan 12 '13 at 17:21
    
Can I do another question and post it up? –  Kaish Jan 12 '13 at 17:37
1  
Sure man, keep em comin –  Alexander Gruber Jan 12 '13 at 17:46
    
@Kaish Updated. –  Alexander Gruber Jan 12 '13 at 20:47
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.