# Construct semi direct product $H \rtimes Q$

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.

• 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

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 $C_7\times C_2$).

• 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
• 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
• Sure man, keep em comin – Alexander Gruber Jan 12 '13 at 17:46
• @Kaish Updated. – Alexander Gruber Jan 12 '13 at 20:47