1) First of all, I have to construct the semi direct product "from H to Q", it is written has $H \lambda Q$ in my notes. I have $H = C_{17}, Q = C_2$.

So I did $Aut(C_{17}) = C_{16}$ and then you can see that 2 divides 16 and so you want to work out how many elements of order 2 there are in $C_{16}$. Doing $\varphi{16} = 2$ and so I have just 2 elements of order 2 in $C_{16}$ and so there are just 2 semi direct products. How do I know what these semi-direct products are? Am I right in assuming that one is the trivial semi direct product, i.e the direct product?

2) Another question, I have to say how many semi direct products there are from H to Q (same notation as before. I have

$H = C_{42}, Q = C_{3}$

So I then get $Aut(H) = C_6 \times C_2$. So from here, I can see that 3 doesn't divide 2 but divides 6 so I just look at that. I want the number of elements of order 3 in $C_6$. I get there are 3 elements of order 3. Assuming $C_6 = \{a, ..., a^6\}$ then the elements of order 3 are $a^2, a^4$ and $a^6$. So from here I get there are three semi direct products for this question. My question is this though, lets assume I had something like $Aut(H) = C_6 \times C_3$, then how would I work this question out? As 3 divides 3 from the other $C_3$ so I can't just ignore it like in the question I had to do. What would I do here? I'm assuming I would have to work out the number of semi direct products in this, and then how would I combine the two?

EDIT: Oh hang on, I need to be able to work out the number of elements in a given order to be able to answer my second question don't I? Dang. Still can't do that :(

  • $\begingroup$ Is $\,\phi 16=2\,$ meant to be "the Euler function of 16 is two"? Because it is completely wrong: a cyclic group of finite order $\,n\,$ has exactly one (cyclic, of course) subgroup of any order $\,d\,$ dividing $\,n\,$ , and thus a cyclic group of even order has one unique subgroup of order two and thus one unique element of order two...od you meant something else? $\endgroup$ – DonAntonio Jan 6 '13 at 21:56
  • $\begingroup$ Perhaps he means that there are two elements satisfying $x^2=1$, which do indeed give rise to the two semidirect products, one of which is a direct product? $\endgroup$ – Derek Holt Jan 6 '13 at 22:00
  • $\begingroup$ @DonAntonio Yeah it was, but I've kind of rushed it when I was typing and mixed everything up in my head, sorry. What I wrote on my paper was this: 2 divides 16 and so we want number of elements of order 2 in $C_{16}$. I.e $ord(a^k) = 2 = \frac{16}{(k, 16)}$. Re arranging this gives us $(k, 16) = 8$ and so we get two elements of order 2 in this semi direct product. The elements or order two that I get (just to show you I understand what I've done) are, if $C_{16} = \{a, a^2, ... , a^{16} \}$, then the two elemnts are $a^8$ and $a^{16}$. I think thats correct. $\endgroup$ – Kaish Jan 6 '13 at 22:03
  • $\begingroup$ No @Kaish, it still isn't correct: $\,a^8\,$ is the only element of order two, whereas $\,a^{16}=1\,$ has order 1. Of course, and as Derek commented, you seem to be interested in elements which fulfil $\,x^2=1\,$. $\endgroup$ – DonAntonio Jan 6 '13 at 22:08

You seem to be seeking semidirect products $\,C_{17}\rtimes C_2\,$ , for which we must know the possible homomorphisms $\,C_2\to\operatorname{Aut}(C_{17})\cong C_{16}\,$ . Since $\,C_{16}\,$ has one single element of order $\,2\,$ there is one unique non-trivial homomorphism as above, which gives the non-abelian group of order $\,34\,$, determined by the action $\,c\cdot a=:a^c=a^{-1}\,\,,\,\,C_2=\langle c\rangle\,\,,\,\,C_{17}=\langle a\rangle \,$ . Thus, we can say that

$$C_{17}\rtimes C_2=\left\{(c^i,a^j)\in C_{17}\times C_2\;\;;\;\;0\leq i\leq 16\,\,,\,0\leq j\leq 1\right\}$$

under the operation


The other homomorphism is the trivial one, which gives us the direct product $\,C_{17}\times C_2\cong C_{34}=$the cyclic group of order $\,34\,$

As for your second question: $\,\operatorname{Aut}(C_{42})\cong C_2\times C_6 ,\,$ so how many homomorphisms $\,C_3\to C_2\times C_6\,$ can you find?...

  • $\begingroup$ How do you get this non abelian group? Where does the $a^c = a6{-1}$ comes from? Also, how do you know under what operation this comes? For your question, like I said, there would be 3 semi direct products. $C_6$ has two elements of order 3, $a^2, a^4$ and one element of order 1, $a^6$. The element of order 1 gives the direct product. The other two semi direct products would be commuting ones as they have the same order. I don't know what they would be though. I don't know the method to show this. $\endgroup$ – Kaish Jan 6 '13 at 22:30
  • $\begingroup$ That was supposed to say, where does the $a^c = a^{-1}$ come from $\endgroup$ – Kaish Jan 6 '13 at 22:36
  • $\begingroup$ It is known that any non-trivial homom. as described in the answer gives a non-abelian semidirect product, but you can prove it directly...An even-order automorphism group of a cyclic group has only one automorphism of order two: the involution (or inversion) automorphism, taking any element to its inverse...I don't understand your question "how do you know under what operation this comes"...The continuation is fine, but the two non-trivial semidirect products are isomorphic, so you get only one (up to isomorphism) non-abelian group in this way, AND the abelian one from the trivial hom. $\endgroup$ – DonAntonio Jan 6 '13 at 22:39
  • $\begingroup$ That operation question was just, you wrote you get that semi direct product (SDP) under that operation. What does that basically mean? So with the other bit, there are two semi direct products, but one of them is non abelian, the other one is and the trivial homomorphism? Sorry, that bit didn't make much sense to me. Also, are they isomorphic as they both have the same order? $\endgroup$ – Kaish Jan 6 '13 at 22:46
  • $\begingroup$ @Kaish, I can't make a complete exposition about semidirect products here. There are excellent books that deal with it, look it up. Just for a taste: $$(c^i,a^j)(c^k,a^r):=(c^{i+(-1)^jk},a^{j+r})$$ since the action is inversion: $$(c^k)^{a^j}=c^{(-1)^jk}$$ as $\,(c^j)^a=c^{-j}\,\,,\,(c^j)^{a^2=1}=c^j\,$ $\endgroup$ – DonAntonio Jan 6 '13 at 22:54

About your first question, you want to construct $$G=N\rtimes H$$ wherein $N=\Bbb Z_{17}=\langle n\rangle, H=\Bbb Z_{2}=\langle h\rangle$. You started correctly because we need some homomorphisms which can construct the proper relations for $G$'s. So you took $$\phi:H\to Aut(N)$$ You noted that correctly that $|Aut(N)|=16$ but we should be careful that any generators of $N$ would be mapped to the any generators of $N$. If we define $$\phi_h, h\in H$$ then it is clear that $\phi_h(n)\in N$. In fact, $\phi_h\in Aut(N)$. So there is for example $l$, $0\leq l\leq 16$ such that $$\phi_h(n)=n^l$$ Note that $H,N$ both are cyclic. Since we assume that $\phi_h$ for all $h\in H$ is an automorphism and $|N|=17$ then we should have $(l,17)=1$. In fact we define $$h^{-1}nh=n^l,\;\;0\leq l\leq 16,\;\;(l,17)=1$$ These are well defined relations for our desired $G$.


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