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I constructed my semidirect product and got the multiplication law to be

$$ab = ab^{64}$$

where as in the answers it said it was

$$ba = ab^8.$$

I want to know if my answer was also correct by asking:

1) Does the order of the LHS matter? So if I wrote $ab = ab^8$, would that have been correct?

2) Is $ab^{64} = ab^8$ as it is just $ab^{8^2}$, which could've been constructed with an element of the same order but differen power (if that makes sense?)

Also, I just thought of this question:I managed to get a nontrivial SDP but if I used Sylow theory, I wouldve only got the trivial one. Here is my question:

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, using Sylow theory, I would split.... ok, I think I've just talked myself through this. I can't use Sylow theory here becuase the group isn't a $p$ group right? I.e I can't split $79$ into primes, $p$ and $q$ to find a $n_p$ and a $n_q$ right, so I just construct my SDP like normal?

EDIT: Let me clarify the Sylow bit. What I would normally use Sylow theory for would be to split my group $H$ into seperate $p$ subgroups. This would show me exactly how many SDP's I have. I.e if I split it into to subgroups, $n_p$ and $n_q$, I can tell that I have 1 SDP as one of these subgroups will be normal and so with two normal groups, I get my SDP to be trivial.

EDIT 2: Also, one final question: Would my final answer for constructing the SDP be the multiplcation law that I get?

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Is b a member of Q? Remember the orders of your elements –  Mr.Guy Jan 12 '13 at 0:58
    
Yeah, and $a$ is in $H$. –  Kaish Jan 12 '13 at 0:59
    
So what is the order of b then? –  Mr.Guy Jan 12 '13 at 1:00
    
Not all groups are commutative: $ab$ and $ba$ are usually very different things. Similarly, the equations $ab=ab^8$ and $ba=ab^8$ are very different. (The former is actually equivalent to $b^7 = 1$) –  Hurkyl Jan 12 '13 at 1:02
    
It is a different equation, but it's very close to the correct one which makes me think you wrote something down wrong or operated incorrectly. –  Mr.Guy Jan 12 '13 at 1:03

1 Answer 1

up vote 1 down vote accepted

Let's use the hint we were given:

If $2$ is of order $39$ in $\Bbb Z_{79}^\ast$ then $8 = 2^3$ has order $\dfrac{39}{(3,39)} = 13$ and thus:

$b \to b^8$ is an automorphism of $\langle b \rangle \cong C_{79}$ of order $13$.

In the semi-direct product $a$ (the element of order $13$) acts on $b$ by conjugation, that is, explicitly, the homomorphism:

$\phi:\langle a \rangle \to \text{Aut}(\langle b \rangle)$ is defined by: $\phi_a(b) = b^8$

If we use the product rule:

$(a,b)*(a',b') = (aa',\phi_{a'}(b)b')$ and identify $\langle a \rangle$ with $\langle a \rangle \times \{e\}$ and $\langle b\rangle$ with $\{e\} \times \langle b \rangle$, we have:

$ba = (e,b)*(a,e) = (a,\phi_a(b)) = (a,b^8) = (a,e)*(e,b^8) = ab^8$

(Note: texts differ on whether the action of conjugation is $b \to a^{-1}ba$ or $b \to aba^{-1}$, so this semi-direct product rule may look slightly different than what you're used to).

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