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 Moderiert von Buri GockelStrukturen und Algebra » Gruppen » Isomorphismen zwischen nicht-zyklischen Gruppen, Erzeuger wird auf Erzeuger abgebildet Druckversion
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 Isomorphismen zwischen nicht-zyklischen Gruppen, Erzeuger wird auf Erzeuger abgebildet
Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Themenstart: 2019-11-17 11:22
 Hi everybody, we are supposed to find all the group $Aut(D_4)$, where $Aut(D_4)$ denotes the set of all automorphisms $D_4\rightarrow D_4$. (i) In the following, suppose we write $D_4$ with the permutation-representation. We first of all showed that if $g$ has order $x$, then $\phi(g)$ must have order $x$ as well. Obviously, $\phi(1)=1$. However, I asked myself whether we don't know more. I mean obviously, I could apply "Brute Force", but I don't fancy this because there are still too many possibilities. But then I found here that it seems as if a generator must be mapped to a generator. In this context, the knowledge of this would simplify things, I guess, because $D_4=\langle\{\left(1 2\right)(3 4)\}, (1 2 3 4)\rangle$. My $$\textbf{question}$$ is the following: Why does it hold that a generator must be mapped to a generator? (I know that there is a proof on StackExchnage, but I cannot understand it because I don't konw what words or alphabets are ... Isn't there another proof for this? Many thanks in advance, Neymar (PS: I don't mind German replies ... :-)

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.1, eingetragen 2019-11-17 11:56
 Hi Neymar, I am not sure if I understand you correctly. Is the following what you want to prove? "Let $G$ and $H$ be groups and let $\varphi: G \to H$ be a surjective homomorphism. Then for every generating set $M \subseteq G$, the image $\varphi(M)$ is a generating subset of $H$". A short proof of this goes as follows: Let $U \subseteq H$ be the subgroup generated by $\varphi(M)$. Then the preimage $\varphi^{-1}(U)$ is a subgroup of $G$ which certainly contains $M$. Hence $\varphi^{-1}(U)=G$ because $M$ generates $G$. This is only possible if $U=H$.

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.2, vom Themenstarter, eingetragen 2019-11-17 15:01
 Hi David, thanks for your reply. Yes, you understood me correctly. :-) My version would have been a bit stronger (let $\phi: G\rightarrow G \in \text{Aut}\left(G\right)$), but apparently, these strong conditions are not required. Since $\phi\left(M\right)\subset G$ (I assume that $H=G$ and that $\phi$ is an automorphism), we then know that each generator must be mapped to a generator, right? (In my next post, I will ask specific questions about the proof.)

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.3, eingetragen 2019-11-17 15:08
 2019-11-17 15:01 - Neymar in Beitrag No. 2 schreibt: Hi David, Since $\phi\left(M\right)\subset G$ (I assume that $H=G$ and that $\phi$ is an automorphism), we then know that each generator must be mapped to a generator, right? What do you mean by "a generator"? We do not obtain $\phi(M)=M$ if $\phi$ is an automorphism or something like that, just that $\phi(M)$ is some subset of $G$ which generates $G$.

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.4, vom Themenstarter, eingetragen 2019-11-17 15:26
 Let's take a concrete example, let's say the dihedral group $D_4$. We defined it in a previous exercise sheet as follows: "Sei $n\geq 3$ und sei $D_n\leq \text{GL}(2,\mathbb R)$ die von der Teilmenge $\left\{ \begin{pmatrix}1&0\\0&-1\end{pmatrix}, \begin{pmatrix}\cos\left( \frac{2\pi}{n}\right)&-\sin\left( \frac{2\pi}{n}\right)\\\sin\left(\frac{2\pi}{n}\right)&\cos\left( \frac{2pi}{n}\right)\end{pmatrix} \right\}\subset \text{GL}(2,\mathbb R)"$ erzeugte Untergruppe." For me, the two matrices $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $\begin{pmatrix}\cos\left( \frac{2\pi}{n}\right)&-\sin\left( \frac{2\pi}{n}\right)\\\sin\left(\frac{2\pi}{n}\right)&\cos\left( \frac{2pi}{n}\right)\end{pmatrix}$ are generators of $D_4$. We are now supposed (in a new exercise) to find ALL automorphisms of $D_4$. I thought I can simplify this by showing (or asking how this can be done) that every automorphism $\phi:G\rightarrow G$ must statisfy $\phi\left(\begin{pmatrix}\cos\left( \frac{2\pi}{n}\right)&-\sin\left( \frac{2\pi}{n}\right)\\\sin\left(\frac{2\pi}{n}\right)&\cos\left( \frac{2pi}{n}\right)\end{pmatrix}\right)=\begin{pmatrix}\cos\left( \frac{2\pi}{n}\right)&-\sin\left( \frac{2\pi}{n}\right)\\\sin\left(\frac{2\pi}{n}\right)&\cos\left( \frac{2pi}{n}\right)\end{pmatrix}$ or $\phi\left(\begin{pmatrix}\cos\left( \frac{2\pi}{n}\right)&-\sin\left( \frac{2\pi}{n}\right)\\\sin\left(\frac{2\pi}{n}\right)&\cos\left( \frac{2pi}{n}\right)\end{pmatrix}\right)=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}$. However, I now realized that the last option cannot be true because the matrix $A:=\begin{pmatrix}1&0\\0 & -1\end{pmatrix}$ has order $2$, i.e., $A^2=\text{id}$, but the rotation matrix has at least order $3$. But is the first option true or not, i.e., it always holds that for any automorphism, $\phi(\text{rotation matrix})=\text{rotation matrix}$? Because if yes, it would be easier to do the task, I guess.

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.5, eingetragen 2019-11-17 15:44
 No, this is not true. For simplicity, I write $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $B=\begin{pmatrix}\cos\left( \frac{2\pi}{n}\right)&-\sin\left( \frac{2\pi}{n}\right)\\\sin\left(\frac{2\pi}{n}\right)&\cos\left( \frac{2pi}{n}\right)\end{pmatrix}$. Let $\phi: D_n \to D_n$ be an automorphism. Then $\{ \phi(A), \phi(B) \}$ is again a set of generators for $D_n$ (as I proved in #1), but this does not mean that $\{ \phi(A), \phi(B) \}=\{A,B \}$. For example, if $n=4$, then $\{ A,B^3\}$ is also a set of generators, so there might be an automorphism $\phi$ with $\phi(A)=A$ and $\phi(B)=B^3$. (I think, there is indeed such an automorphism, but at the moment I am not completely sure about this)

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.6, vom Themenstarter, eingetragen 2019-11-17 16:13
 Oh, thanks for the answer! How would you approach the task then? Do you think that I can somehow use what you proved in "Beitrag 1"? If it is okay, I will now use another notation for the group $D_4$: $D_4 = \{(1), (2 4), (1 3), (1 2)(3 4), (1 4)(2 3), (1 3)(2 4), (1 2 3 4), (1 4 3 2)\}.$ There is one element of order $4$: $(1)$. There are two elements of order $4$: $(1 2 3 4), (1 4 3 2)$. The rest of the elements all have order $2$. I hope you see why I don't want to apply Brute Force: There are simply too many possibilities ... (I just got an idea 10 mintues ago: Using the conjugacy classes, since if two elements $a,b$ are conjugate to each other, then this must hold for their images as well. I now there are only 4 conjugacy classes, I guess I will try to use it now.)

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.7, eingetragen 2019-11-17 17:30
 I suggest the following approach: We should first fix a generating set for $D_4$, for example $\sigma=(12)$ and $\rho=(1234)$. Now an automorphism $\phi$ is uniquely determined by the images of $\sigma$ and $\rho$. Since $\rho$ has order 4, $\phi(\rho)$ also has order 4, so there are only two possibilities for $\phi(\rho)$. Can you go on from this point?

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.8, vom Themenstarter, eingetragen 2019-11-17 18:19
 Not yet. :-) Let's fix as the two generating elements $\sigma = \left(1 4\right)\left( 2 3 \right)$ and $\rho=\left(1 2 3 4\right)$, where I have used other elements than you had suggested (because I think that you mistyped the elements). Now $\phi(\rho)=(1 2 3 4)$ or $\phi(1 2 3 4)=(1 4 3 2)$. But for $\sigma$, there are $5$ possibilities. (This is sth we knew prior to choosing a set of generators, so why is it fruitful to choose generators? Maybe I don't know yet what I means that a homomorphism is uniquely determined by the images of its generators.)

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.9, eingetragen 2019-11-17 18:34
 2019-11-17 18:19 - Neymar in Beitrag No. 8 schreibt: Let's fix as the two generating elements $\sigma = \left(1 4\right)\left( 2 3 \right)$ and $\rho=\left(1 2 3 4\right)$, where I have used other elements than you had suggested (because I think that you mistyped the elements). You are right, I had a mistake there. Indeed we have 2 possibilities for $\phi(\rho)$ and 5 for $\phi(\sigma)$. Now, since $\{ \rho, \sigma \}$ generates $D_4$, once we know $\phi(\rho)$ and $\phi(\sigma)$, we know the whole automorphism $\phi$, if there is such an automorphism. So there can be at most $2 \cdot 5=10$ automorphisms. However, it is not clear (and indeed not true here), that for every possible choice of $\phi(\rho)$ and $\phi(\sigma)$ there is indeed an automorphism, so you still need to check which of these 10 possibilities really give an automorphism. For this it is useful to know the following: every element of $D_4$ can be written in the form $\sigma^i \rho^j$ with $i \in \{0,1 \}$ and $j \in \{ 0,1,2,3 \}$.

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.10, vom Themenstarter, eingetragen 2019-11-17 18:44
 Wait a minute. :-) Assume we chose  $\phi(\rho)=\left(1 2 3 4\right)$ and $\phi(\sigma)=\sigma=(1 4)(2 3)$, where would I know from what e.g. $\phi(1 3)$ is? I think it would really help me if I understood in this example. And yeah, this is exactly what I was searching for: I was searching for some way to reduce the huge number of possibilities to some few.

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.11, eingetragen 2019-11-17 19:00
 We have $(13)=\sigma \rho$, so $\phi(13)=\phi(\sigma) \phi(\rho)$. In general, every element of a group $G$ generated by some set $M$ can be written as a product of the form $\alpha_1 \cdots \alpha_n$ where for each $i$, we have $\alpha_i \in M$ or $\alpha_i^{-1} \in M$, so with the same argument, once we know the image of every element of $M$ under some homomorphism $\phi$, we know the image of any element of $G$ under $\phi$.

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.12, vom Themenstarter, eingetragen 2019-11-17 21:13
 Oh yes, thanks for the hint. I have tried for one case and wanted to see whether I am on the right path or not. :-) We know that any $x\in D_4$ can be written as a product of $\rho^i\sigma^j$. $$\textbf{Case 1.}$$ $\phi(p)=\rho, \phi(\sigma)=(2 4) [= \rho\sigma]$. $\phi\left(\left(2 4\right)\right)=\dots = \left( 1 2\right)(3 4), \phi\left(\left(1 3\right)\right)=\phi(\sigma\rho)=(1 4)(2 3), \phi\left(\left(1 2\right)\left( 3 4\right)\right) = (1 3), \phi\left(\left(1 3\right)\left( 2 4\right)\right)=\rho^2=(1 3)(2 4), \phi\left(\left(1 4 3 2\right)\right) = \phi(\rho^{-1})=\phi(\rho)^{-1}$.   I hope that I didn't mistype any permutations. (Obiously, $\phi(1)=1$.) How can I tell whether this automorphism is allowed or not? (Here, it doesn't look like any rules were broken.) PS: When it comes to automorphisms, we always know that one exists, which is the trivial one.

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.13, eingetragen 2019-11-17 21:32
 2019-11-17 21:13 - Neymar in Beitrag No. 12 schreibt: $$\textbf{Case 1.}$$ $\phi(p)=\rho, \phi(\sigma)=(2 4) [= \rho\sigma]$. $\phi\left(\left(2 4\right)\right)=\dots = \left( 1 2\right)(3 4), \phi\left(\left(1 3\right)\right)=\phi(\sigma\rho)=(1 4)(2 3), \phi\left(\left(1 2\right)\left( 3 4\right)\right) = (1 3), \phi\left(\left(1 3\right)\left( 2 4\right)\right)=\rho^2=(1 3)(2 4), \phi\left(\left(1 4 3 2\right)\right) = \phi(\rho^{-1})=\phi(\rho)^{-1}$.   Correct. As I said in #10 we can write every element of $D_4$ in the form $\sigma^i \rho^j$ with $i \in \{0,1 \}$ and $j \in \{0,1,2,3\}$. So once we fixed $\phi(\sigma)$ and $\phi(\rho)$, we can write $\phi$ as $\sigma^i \rho^j \mapsto \phi(\sigma)^i \phi(\rho)^j.$ Now we need to check for which of the ten possibilities this map is indeed an automorphism. So you need to check whether they are homomorphisms and whether they are bijective. (I probably won't have time to write anything more here before tomorrow evening.)

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.14, vom Themenstarter, eingetragen 2019-11-18 19:59
 Hi DavidM, I just managed to go through all ten possibilities. :-) Unfortunately, there were only 2 cases where bijectivity was not satisfied, therefore I obtain 8 bijective maps. (Unfortunately, because realizing that bijectivity was not satisfied accelerated the process.) For some maps, I tried with one example to check whether homomorphism might not be satisfied because I don't feel like checking that $\forall x,y\in D_4: \phi(xy)=\phi(x)\phi(y)$. Done properly, this would take too much time. What I realized is the following though: We never used that $(1 2 3 4)$ and $(1 4 3 2)$ are conjugate to each other (another example: $(1 2)(3 4)$ and $(1 4)(2 3)$ are conjugate to each other). Actually, this was (a). (b): I think I already showed that $\text{Inn(G)}\trianglelefteq \text{Aut}(G)$. So for me it remains to determine $\text{Out}(D_4)$. By definition, the factor group (or quotient group) $\text{Out}(D_4)$ consists of all left cosets, i.e., $\text{Out}\left(D_4\right)=\left\{ \phi\text{Inn}(Q_8) \mid \phi\in\text{Aut}(G)\right\}.$ Ergo, I thought that I first need to determine $\text{Inn}(G)$. I thought a bit of how I can do this and the best I came up with is to determine $\gamma_g$ for all $g\in G$, which can be done by brute force. I actually started and have the suspicion that $\gamma_{(1 2 3 4)}=\gamma_{(1 4 3 2)}$. For all the other elements it holds that $g=g^{-1}$, so for these elements I don't have any better idea than explicit calculation ... I guess that $\text{Inn}(Q_8)$ has 7 elements. Then I would determine $\text{Out}(D_4)$ via explicit calculation as well. --Neymar

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.15, eingetragen 2019-11-18 20:28
 2019-11-18 19:59 - Neymar in Beitrag No. 14 schreibt: Hi DavidM, I just managed to go through all ten possibilities. :-) Unfortunately, there were only 2 cases where bijectivity was not satisfied, therefore I obtain 8 bijective maps. (Unfortunately, because realizing that bijectivity was not satisfied accelerated the process.) For some maps, I tried with one example to check whether homomorphism might not be satisfied because I don't feel like checking that $\forall x,y\in D_4: \phi(xy)=\phi(x)\phi(y)$. Done properly, this would take too much time. First of all: you are right, 8 of the 10 maps are bijective. In fact, all of these eight maps are automorphisms. At the moment, I am not sure if there is a really short and elementary argument to see this - perhaps there is one, I am not an expert in this. What I realized is the following though: We never used that $(1 2 3 4)$ and $(1 4 3 2)$ are conjugate to each other (another example: $(1 2)(3 4)$ and $(1 4)(2 3)$ are conjugate to each other). Did anyone say that we should use this? At the moment, I do not really see how - as I said, I am not an expert, so it is possible that it is useful. Actually, this was (a). (b): I think I already showed that $\text{Inn(G)}\trianglelefteq \text{Aut}(G)$. So for me it remains to determine $\text{Out}(D_4)$. By definition, the factor group (or quotient group) $\text{Out}(D_4)$ consists of all left cosets, i.e., $\text{Out}\left(D_4\right)=\left\{ \phi\text{Inn}(Q_8) \mid \phi\in\text{Aut}(G)\right\}.$ Ergo, I thought that I first need to determine $\text{Inn}(G)$. I thought a bit of how I can do this and the best I came up with is to determine $\gamma_g$ for all $g\in G$, which can be done by brute force. I actually started and have the suspicion that $\gamma_{(1 2 3 4)}=\gamma_{(1 4 3 2)}$. For all the other elements it holds that $g=g^{-1}$, so for these elements I don't have any better idea than explicit calculation ... I guess that $\text{Inn}(Q_8)$ has 7 elements. Then I would determine $\text{Out}(D_4)$ via explicit calculation as well. --Neymar $|\mathrm{Inn}(Q_8)|=7$ cannot be true. In your other thread you have already seen that $\mathrm{Aut}(Q_8) \cong S_4$, so $|\mathrm{Aut}(Q_8)|=24$. Now, $\mathrm{Inn}(Q_8)$ is a subgroup of $\mathrm{Aut}(Q_8)$, so its order must be a divisor of 24 and therefore cannot be 7. For the inner automorphisms, it is useful to know that the kernel of the map $\gamma$ mentioned in the task is the center of $G$ (can be proved easily). Therefore $|\mathrm{Inn}(G)|=\frac{|G|}{|Z(G)|}$ where I wrote $Z(G)$ for the center of $G$.

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.16, vom Themenstarter, eingetragen 2019-11-18 22:08
 Nice argumentation. :-) This would then imply that $|\text{Inn}(Q_8)|=4$. I am still with you that $\text{ker}\left(\Phi\right)=Z(Q_8)=\{\pm 1\}$. However, I do not see yet how we get to your last equation.

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.17, vom Themenstarter, eingetragen 2019-11-18 22:35
 Oh, I am so sorry! I mixed up $D_4$ and $Q_8$. Alright, let us find $\text{Inn}(\mathbf D_4)$. We know that $\left|\text{Aut}\left(D_4\right)\right|=8$. Therefore, for $Inn(D_4)$, there are not so many possibilities: $|Inn(D_4)|\in\{1,2,4,8\}$. EDIT: But since we are supposed to find $Inn(Q_8)$ as well, I wouldn't mind hitting two birds with one stone. :-)

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.18, eingetragen 2019-11-19 18:23
 2019-11-18 22:08 - Neymar in Beitrag No. 16 schreibt: Nice argumentation. :-) This would then imply that $|\text{Inn}(Q_8)|=4$. I am still with you that $\text{ker}\left(\Phi\right)=Z(Q_8)=\{\pm 1\}$. Correct. However, I do not see yet how we get to your last equation. You are confusing me a bit with your notation. I assume by $\Phi$ you mean what was called $\gamma$ before. I will continue calling it $\gamma$. Then we of course have $G/\ker(\gamma) \cong \mathrm{im}(\gamma)$, so $|\mathrm{im}(\gamma)|=\frac{|G|}{|\ker(\gamma)|}$. Now with $\ker(\gamma)=Z(G)$ and $\mathrm{im}(\gamma)=\mathrm{Inn}(G)$. 2019-11-18 22:35 - Neymar in Beitrag No. 17 schreibt: Oh, I am so sorry! I mixed up $D_4$ and $Q_8$. Alright, let us find $\text{Inn}(\mathbf D_4)$. We know that $\left|\text{Aut}\left(D_4\right)\right|=8$. Therefore, for $Inn(D_4)$, there are not so many possibilities: $|Inn(D_4)|\in\{1,2,4,8\}$. Do you know what the center of $D_4$ is?

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.19, vom Themenstarter, eingetragen 2019-11-20 01:26
 Yes, the center of $D_4$ is $\{(1), (1 3)(2 4)\}$. Thanks!

Neymar
Aktiv
Dabei seit: 03.01.2019
Mitteilungen: 663
 Beitrag No.20, vom Themenstarter, eingetragen 2019-11-20 02:28
 $G\ker(\gamma)\cong im(\gamma)$ Could you elaborate a bit further on this, please?

DavidM
Senior
Dabei seit: 11.06.2012
Mitteilungen: 265
 Beitrag No.21, eingetragen 2019-11-20 18:41
 2019-11-20 02:28 - Neymar in Beitrag No. 20 schreibt: $G\ker(\gamma)\cong im(\gamma)$ Could you elaborate a bit further on this, please? For any homomorphism of groups $\varphi: G \to H$ there is an isomorphism $G/\ker(\varphi) \cong \mathrm{im}(\varphi)$. This is a basic theorem in group theory. In the german literature it is called "Homomorphiesatz". I am not sure how it is usually called in English, with a quick search I found "fundamental theorem on homomorphisms" or "first isomorphism theorem".

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