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Construct explcitly a non-commutative semi direct product $H \rtimes Q$ with

$$H = C_{79} \hspace{2cm} Q = C_{13}.$$

You may assume that the least positive integer $k \geq 1$ such that $2^k \equiv 1 \mod 79$ is $k = 39$.

So I do everything normal to get my SDP. Let $H = \langle b \rangle, Q = \langle a \rangle$ and $\mu$ generate $\mathrm{Aut}(H)$. I have the map $\mu: b \mapsto b^d$ and I got my $d$ to be $2$ as gcd$(2, 79) = 1$. Now I want to find an element, $r$ of $\mu$ (so in $C_{78}$) which has order $13$ and so I can map

$$\mu^r : b \mapsto b^{2^r}$$

and this will give me my multiplication order. So, I got one element of order $13$ to be $\mu^6$, meaning $r = 6$.

$2^6 = 64$. So now I get my multiplication order to be

$$ba = ab^{64}$$

But in the answers, it says it should $ba = ab^8$ as I should've got $r = 3$, but from my method to work out the value of $r$, the lowest value I get for it is $6$. Why is $r$ supposed to equal $3$?

EDIT: If I've rushed any description or you don't understand one bit I've written, then please ask.

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1 Answer 1

up vote 1 down vote accepted

You are given that $2$ has order $39=3\cdot 13$ and need an element of order $13$. Then $2^3=8$ suggests itself.

On the other hand, $64$ is fine as well. Remember that you merely need that $ba^{13}$ turns out as $b$ after shifting all those over: In $$b=ba^{13}=ab^ra^{12}=a^2b^{r^2}a^{11}=\ldots =a^{13}b^{r^{13}}$$ we merely need that $r^{13}\equiv 1\pmod {79}$. In fact, since $x^{78}\equiv 1\pmod{79}$ for any nonzero $x$ and $78=6\cdot 13$, you could choose $r=x^6$ for arbitrary nonzero $x$. You took $x=2$ and then $r=64$, which is fine. One might as well have picked $x=9$ and then $r=531441\equiv 8\pmod{79}$. All choices are fine as long as $r\not\equiv 1$ (as we want a non-abelian group).

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That's what it says the answers, how do you know that $2$ has order $39$? –  Kaish Jan 13 '13 at 18:38
    
Also, lets say I missed that, is my multiplication order wrong them? –  Kaish Jan 13 '13 at 18:39

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