Aut(D_12) Elements & Inner Automorphisms: Dihedral Group with 12 Elements

[Mr Davis 97]

Homework Statement

Find all the elements of $\operatorname{Aut}(D_{12})$. Which elements are inner automorphisms?

Note: $D_{12}$ is the dihedral group with $12$ elements.

Homework Equations

The Attempt at a Solution

I'm not really sure how to approach this. What I do know is that $|D_{12}/Z(D_{12})| = 12/2 = 6$, so $|\operatorname{Inn}(D_{12})| = 6$. So $6 \le |\operatorname{Aut}(D_{12})|$, but I don't have an upper bound, and this also hasn't helped me explicitly find what the automorphisms are.

 

[fresh_42]

Homework Statement

Find all the elements of $\operatorname{Aut}(D_{12})$. Which elements are inner automorphisms?

Note: $D_{12}$ is the dihedral group with $12$ elements.

Homework Equations

The Attempt at a Solution

I'm not really sure how to approach this. What I do know is that $|D_{12}/Z(D_{12})| = 12/2 = 6$, so $|\operatorname{Inn}(D_{12})| = 6$. So $6 \le |\operatorname{Aut}(D_1{2})|$, but I don't have an upper bound, and this also hasn't helped me explicitly find what the automorphisms are.

You know a bit more. There are only two groups of order $6$, so you can probably determine $\operatorname{Inn}(D_{12})\,,$ i.e. whether it is $\mathbb{Z}_6$ or $S_3$. I would start to check where repeated conjugations with $sr$ lead to. We also have $\operatorname{Inn}(D_{12}) \trianglelefteq \operatorname{Aut}(D_{12})$, so we have multiples of $6$ as the order of the automorphism group. You could start to play a bit with the three relations: $r^6=1\; , \;s^2=1\; , \;srsr=1$ and see whether you can derive restrictions on a potential automorphism $\sigma$. Maybe it also helps to calculate modulo $\operatorname{Inn}(D_{12})$, but that's just an idea - I haven't checked it.

Example: Let's assume $\sigma \in \operatorname{Aut}(D_{12}) - \{\,1\,\}$. Then $\sigma (s) \in \{\, r^3\in Z(D_{12})\; , \;sr^n\,\}$.

 

[Mr Davis 97]

You know a bit more. There are only two groups of order $6$, so you can probably determine $\operatorname{Inn}(D_{12})\,,$ i.e. whether it is $\mathbb{Z}_6$ or $S_3$. I would start to check where repeated conjugations with $sr$ lead to. We also have $\operatorname{Inn}(D_{12}) \trianglelefteq \operatorname{Aut}(D_{12})$, so we have multiples of $6$ as the order of the automorphism group. You could start to play a bit with the three relations: $r^6=1\; , \;s^2=1\; , \;srsr=1$ and see whether you can derive restrictions on a potential automorphism $\sigma$. Maybe it also helps to calculate modulo $\operatorname{Inn}(D_{12})$, but that's just an idea - I haven't checked it.

Example: Let's assume $\sigma \in \operatorname{Aut}(D_{12}) - \{\,1\,\}$. Then $\sigma (s) \in \{\, r^3\in Z(D_{12})\; , \;sr^n\,\}$.

I think I need a bit more help on this one. When I eventually list the automorphisms, as the problem says to, how will I ever go about listing them? One time I listed the elements in $\operatorname{Aut}(\mathbb{Z}/4\mathbb{Z})$, and I did this my explicitly listing out which elements mapped to which elements in each automorphism; however, this was easy since there were so few elements. $D_{12}$ has 12, so this seems much more difficult.

 

[fresh_42]

Well, I haven't done the exercise, only read the solution. Let's see what we have and assume $\sigma \in \operatorname{Aut}(D_{12}) -\{\,1\,\}$.

Then $\sigma(s)$ is of order $2$ and all elements of order two are $s,r^3,sr^n$, if I made no mistake. We also know $\sigma(Z(D_{12}))=Z(D_{12})=\{\,1,r^3\,\}$. Since we can rule out $1$ as image, we have $\sigma(r^3)=r^3$ or $r^{-1}\cdot \sigma(r) \in \{\,1,r^2,r^4\,\}$ as only elements of order three. Then $\sigma(r) \in \{\,r,r^3,r^5\,\}$. So we have all potential images of the generators, which are maximal $7\cdot 3$ possibilities. I assume that $\sigma(r)=r^3$ leads to a contradiction, so with $6\,|\,|\operatorname{Aut}(D_{12})|$ and $6 \leq |\operatorname{Aut}(D_{12})| \leq 21$ we have only left, that either there are no outer automorphisms (there are!), or $|\operatorname{Aut}(D_{12})|\in \{\,12,18\,\}$.

The combination $(r,s) \longmapsto (r,s)$ is the identity and $(r,s) \longmapsto (r^k,r^3)$ are impossible. So in a similar way, you can rule out combinations. You could start to determine all $6$ inner automorphisms.

 

[Mr Davis 97]

Well, I haven't done the exercise, only read the solution. Let's see what we have and assume $\sigma \in \operatorname{Aut}(D_{12}) -\{\,1\,\}$.

Then $\sigma(s)$ is of order $2$ and all elements of order two are $s,r^3,sr^n$, if I made no mistake. We also know $\sigma(Z(D_{12}))=Z(D_{12})=\{\,1,r^3\,\}$. Since we can rule out $1$ as image, we have $\sigma(r^3)=r^3$ or $r^{-1}\cdot \sigma(r) \in \{\,1,r^2,r^4\,\}$ as only elements of order three. Then $\sigma(r) \in \{\,r,r^3,r^5\,\}$. So we have all potential images of the generators, which are maximal $7\cdot 3$ possibilities. I assume that $\sigma(r)=r^3$ leads to a contradiction, so with $6\,|\,|\operatorname{Aut}(D_{12})|$ and $6 \leq |\operatorname{Aut}(D_{12})| \leq 21$ we have only left, that either there are no outer automorphisms (there are!), or $|\operatorname{Aut}(D_{12})|\in \{\,12,18\,\}$.

The combination $(r,s) \longmapsto (r,s)$ is the identity and $(r,s) \longmapsto (r^k,r^3)$ are impossible. So in a similar way, you can rule out combinations. You could start to determine all $6$ inner automorphisms.

Quick question. Why do I have to only look at where the generators go? I that there is a statement called the Universal Mapping Property for presentations, and this says that if you have the generators mapping to elements in the codomain that satisfy the same relation, then there is a unique homomorphism. But why can't there be homomorphisms not described in this way?

 

[fresh_42]

Quick question. Why do I have to only look at where the generators go?

Because the rest is then determined via the homomorphism property: $\sigma(r^ns^mr^k\ldots)=\sigma(r)^n\sigma(s)^m\sigma(r)^k\ldots$

I that there is a statement called the Universal Mapping Property for presentations, and this says that if you have the generators mapping to elements in the codomain that satisfy the same relation, then there is a unique homomorphism. But why can't there be homomorphisms not described in this way?

There can always be a homomorphism not described that way. But all homomorphisms are determined by their images of the generators plus the requirement to be a homomorphism plus the need, that relations are preserved, as we need $\sigma(1)=1$. Not sure I understood your question correctly.

 

[Mr Davis 97]

Because the rest is then determined via the homomorphism property: $\sigma(r^ns^mr^k\ldots)=\sigma(r)^n\sigma(s)^m\sigma(r)^k\ldots$

There can always be a homomorphism not described that way. But all homomorphisms are determined by their images of the generators plus the requirement to be a homomorphism plus the need, that relations are preserved, as we need $\sigma(1)=1$. Not sure I understood your question correctly.

I guess maybe my question can be answer by whether you think this proof is correct or not:

We start by getting bounds on $|\operatorname{Aut}(D_{12})|$. We know that $Z(D_{12}) = \{1,r^3\}$. So $$|D_{12}/Z(D_{12})| = 12/2 = 6 \le |\operatorname{Inn}(G)|.$$ But $\operatorname{Inn}(D_{12})\le\operatorname{Aut}(D_{12})$, so $6\le |\operatorname{Aut}(D_{12})|$, and we have a lower bound. Also, by Lagrange's theorem, $|\le\operatorname{Aut}(D_{12})|$ must be a multiple of 6.

Next, we will try to find an upper bound by looking at isomorphism invariants to see what automorphisms are possible.
Let $\pi\in\operatorname{Aut}(D_{12})$. We will look at where $\pi$ takes the generators. Since for $D_{12}$ there is only one subgroup of order $6$, namely $\langle r \rangle$, we have to have that whatever $r$ maps to must generate a group of order $6$ as well. So we see that $r \mapsto r^k$ where $\gcd(k,6)=1$ as the possibilities. This gives $2$ possibilities: $\pi(r) \in \{r,r^5\}$. Now we look at where $s$ must go. Note that order is an isomorphism invariant, so since $|s|=2$, the order of the element $s$ maps to must also be $2$. So clearly $s\mapsto sr^l$ where $0 \le l < 6$, since all of these are reflections and hence have order $2$. Note that the only other element of order $2$ is $r^3$, but if $s\mapsto r^3$, then we clearly wouldn't have a bijection, since $\pi$ could never map to any of the reflections. Hence $\pi(s)\in \{s,sr,sr^2,sr^3,sr^4,sr^5\}$. So we have $2\cdot 6 =12$ possibilities for automorphisms. So $|\operatorname{Aut}(D_{12})| = 6\text{ or }12$

We will show that we indeed have $12$ automorphisms. First, note that $\pi$ is a bijection. This is because if $l\in[0,6)$ and $k\in\{1,5\}$, then $\langle r^k, sr^l \rangle = D_{12}$. So we have a surjective map from one finite set to the same finite set, which indicates that we have a bijection. Second, note that $\pi$ is a homomorphism. This is because it preserves multiplication. We can prove this by showing that the same relations are satisfied by $\pi(r)$ and $\pi(s)$:
$\begin{align} [\pi(r)]^6 &= (r^k)^6=(r^6)^k = 1^k = 1\\ [\pi(s)]^2 &= sr^lsr^l = ssr^{-l}r^l = q\\ [\pi(s)\pi(r)\pi(s)] &= sr^lr^ksr^l = ssr^{-l-k}r^l = r^{-k} = [\pi(r)]^{-1} \end{align}$

Hence, we have shown that there are $12$ automorphisms and we have described precisely what they are.

 

[fresh_42]

I guess maybe my question can be answer by whether you think this proof is correct or not:

We start by getting bounds on $|\operatorname{Aut}(D_{12})|$. We know that $Z(D_{12}) = \{1,r^3\}$. So $$|D_{12}/Z(D_{12})| = 12/2 = 6 \le |\operatorname{Inn}(G)|.$$ But $\operatorname{Inn}(D_{12})\le\operatorname{Aut}(D_{12})$, so $6\le |\operatorname{Aut}(D_{12})|$, and we have a lower bound. Also, by Lagrange's theorem, $|\le\operatorname{Aut}(D_{12})|$ must be a multiple of 6.

Next, we will try to find an upper bound by looking at isomorphism invariants to see what automorphisms are possible.
Let $\pi\in\operatorname{Aut}(D_{12})$. We will look at where $\pi$ takes the generators. Since for $D_{12}$ there is only one subgroup of order $6$, namely $\langle r \rangle$, we have to have that whatever $r$ maps to must generate a group of order $6$ as well. So we see that $r \mapsto r^k$ where $\gcd(k,6)=1$ as the possibilities. This gives $2$ possibilities: $\pi(r) \in \{r,r^5\}$. Now we look at where $s$ must go. Note that order is an isomorphism invariant, so since $|s|=2$, the order of the element $s$ maps to must also be $2$. So clearly $s\mapsto sr^l$ where $0 \le l < 6$, since all of these are reflections and hence have order $2$. Note that the only other element of order $2$ is $r^3$, but if $s\mapsto r^3$, then we clearly wouldn't have a bijection, since $\pi$ could never map to any of the reflections. Hence $\pi(s)\in \{s,sr,sr^2,sr^3,sr^4,sr^5\}$. So we have $2\cdot 6 =12$ possibilities for automorphisms. So $|\operatorname{Aut}(D_{12})| = 6\text{ or }12$

We will show that we indeed have $12$ automorphisms. First, note that ...

[Perfect up to here. $(*)$.]
... all possible combinations for

... $\pi$ is

are all

a bijection.

With only $\pi$ a bijection, it's not clear what you meant, as we started with the assumption that it is one.

This is because if $l\in[0,6)$ and $k\in\{1,5\}$, then $\langle r^k, sr^l \rangle = D_{12}$.

We note that $r^5=r^{-1}$, so $s=(sr^l)\cdot r^{-l}$ and we get the entire group.

So we have a surjective map from one finite set to the same finite set, which indicates that we have a bijection.

Now you lost me. We have $12$ combinations $(l,k)$ and each is a bijection on $D_{12}$ and we have $|\operatorname{Aut}(D_{12})|\in \{\,6,12\,\}$. How does this rule out, that two different combinations represent the same homomorphism?

Second, note that $\pi$ is a homomorphism.

We started off by assuming this. Why do we need to show it now? We required it. My confusion already started at $(*)$ as you can see from my previous remarks. I think you should have distinguished between the given automorphism $\pi$ at the beginning which you used to derive necessary conditions, and a potential automorphism $\hat{\pi}$ defined by $(l,k)$, which starts as a mapping and we must show that it is an automorphism, plus that they all are different.

I suppose you wanted to reason along the following lines:
Given any combination $(l,k)$ with the restrictions above, then $\hat{\pi}(r):=r^k\; , \;\hat{\pi}(s):=sr^l$ defines an automorphism.
Proof:
$\hat{\pi}$ are all bijiective, see above.
$\hat{\pi}$ are defined on the generators and they preserve the relations, see below.
So there is a unique homomorphism $\pi$ which extends $\hat{\pi}$ by the universal property.

Here is a little flaw. For the bijection argument, at least as I understood it, you already used the homomorphism property (in order to reach all elements of $D_{12}$) but you do not have it yet. Can we either find another argument, or can we first deduce the extension form $\hat{\pi}$ to a homomorphism $\pi$ and show bijectivity last?

This is because it preserves multiplication. We can prove this by showing that the same relations are satisfied by $\pi(r)$ and $\pi(s)$:
$\begin{align} [\pi(r)]^6 &= (r^k)^6=(r^6)^k = 1^k = 1\\ [\pi(s)]^2 &= sr^lsr^l = ssr^{-l}r^l = q\\ [\pi(s)\pi(r)\pi(s)] &= sr^lr^ksr^l = ssr^{-l-k}r^l = r^{-k} = [\pi(r)]^{-1} \end{align}$

Hence, we have shown that there are $12$ automorphisms and we have described precisely what they are.

Remains to show, that they are all different. We do have a surjection from $S:=\{\,(l,k)\,|\,\ldots\,\}$ onto $\operatorname{Aut}(D_{12})$ with $|S|=12$ and $|\operatorname{Aut}(D_{12})| \in \{\,6,12\,\},$ but why is it also injective? It could (theoretically) be, that we map two on one, because we don't know $\operatorname{Inn}(D_{12}) \lneq \operatorname{Aut}(D_{12})$ by now. Maybe you should list all inner automorphisms first, as I thought it is part of the exercise anyway.

 

[Mr Davis 97]

[Perfect up to here. $(*)$.]
... all possible combinations for

are all

With only $\pi$ a bijection, it's not clear what you meant, as we started with the assumption that it is one.

We note that $r^5=r^{-1}$, so $s=(sr^l)\cdot r^{-l}$ and we get the entire group.

Now you lost me. We have $12$ combinations $(l,k)$ and each is a bijection on $D_{12}$ and we have $|\operatorname{Aut}(D_{12})|\in \{\,6,12\,\}$. How does this rule out, that two different combinations represent the same homomorphism?

We started off by assuming this. Why do we need to show it now? We required it. My confusion already started at $(*)$ as you can see from my previous remarks. I think you should have distinguished between the given automorphism $\pi$ at the beginning which you used to derive necessary conditions, and a potential automorphism $\hat{\pi}$ defined by $(l,k)$, which starts as a mapping and we must show that it is an automorphism, plus that they all are different.

I suppose you wanted to reason along the following lines:
Given any combination $(l,k)$ with the restrictions above, then $\hat{\pi}(r):=r^k\; , \;\hat{\pi}(s):=sr^l$ defines an automorphism.
Proof:
$\hat{\pi}$ are all bijiective, see above.
$\hat{\pi}$ are defined on the generators and they preserve the relations, see below.
So there is a unique homomorphism $\pi$ which extends $\hat{\pi}$ by the universal property.

Here is a little flaw. For the bijection argument, at least as I understood it, you already used the homomorphism property (in order to reach all elements of $D_{12}$) but you do not have it yet. Can we either find another argument, or can we first deduce the extension form $\hat{\pi}$ to a homomorphism $\pi$ and show bijectivity last?

Remains to show, that they are all different. We do have a surjection from $S:=\{\,(l,k)\,|\,\ldots\,\}$ onto $\operatorname{Aut}(D_{12})$ with $|S|=12$ and $|\operatorname{Aut}(D_{12})| \in \{\,6,12\,\},$ but why is it also injective? It could (theoretically) be, that we map two on one, because we don't know $\operatorname{Inn}(D_{12}) \lneq \operatorname{Aut}(D_{12})$ by now. Maybe you should list all inner automorphisms first, as I thought it is part of the exercise anyway.

Okay, I took all of your comments into account. So here is where I am now. I have shown that there are at most 12 automorphisms and I have described what they are. But $|\operatorname{Inn}(D_{12})| = 6$ and $|\operatorname{Aut}(D_{12})| = 6\text{ or }12$. So if I can find an automorphism that is not inner, then I would be done in showing that $|\operatorname{Aut}(D_{12})| = 12$?, right So how can I find an automorphism that is not inner?

 

[fresh_42]

Okay, I took all of your comments into account. So here is where I am now. I have shown that there are at most 12 automorphisms and I have described what they are. But $|\operatorname{Inn}(D_{12})| = 6$ and $|\operatorname{Aut}(D_{12})| = 6\text{ or }12$. So if I can find an automorphism that is not inner, then I would be done in showing that $|\operatorname{Aut}(D_{12})| = 12$?, right So how can I find an automorphism that is not inner?

My suspicion is, that $(l,k)=(l,1)$ are inner and $(l,k)=(l,-1)$ are outer, or vice versa. But to show $\pi(l,k) = \pi(n,m) \Longrightarrow l=n,k=m$ should be easy as well.

 

[Mr Davis 97]

My suspicion is, that $(l,k)=(l,1)$ are inner and $(l,k)=(l,-1)$ are outer, or vice versa. But to show $\pi(l,k) = \pi(n,m) \Longrightarrow l=n,k=m$ should be easy as well.

Does this work?

Let $\phi$ be the automorphism such that $\phi(r)=r$ and $\phi(s)=sr$. Suppose for contradiction that $\phi_g(x) = gxg^{-1}$ for some $g\in D_{12}$. Then $\phi_g(r) = grg^{-1}=r$ and $\phi_g(s) = gsg^{-1} = sr$. So then $gr=rg$ and $gs = srg$, which both taken together imply that $gs = sgr$. Let $g = s^ir^j$. The case when $i=1$ immediately reduces to the $i=0$ case, so we take the latter case: $$r^js=sr^jr \implies sr^{-j} = sr^{j+1} \implies r^{-j} = r^{j+1} \implies r = 1.$$ But this is a contradiction. So $\phi$ is an automorphism that is not inner, and so $|\operatorname{Aut}(D_{12})| \ge 7$, so $|\operatorname{Aut}(D_{12})| = 12$.

 

[fresh_42]

Does this work?

Let $\phi$ be the automorphism such that $\phi(r)=r$ and $\phi(s)=sr$. Suppose for contradiction that $\phi_g(x) = gxg^{-1}$ for some $g\in D_{12}$. Then $\phi_g(r) = grg^{-1}=r$ and $\phi_g(s) = gsg^{-1} = sr$. So then $gr=rg$ and $gs = srg$, which both taken together imply that $gs = sgr$. Let $g = s^ir^j$. The case when $i=1$ immediately reduces to the $i=0$ case, so we take the latter case: $$r^js=sr^jr \implies sr^{-j} = sr^{j+1} \implies r^{-j} = r^{j+1} \implies r = 1.$$ But this is a contradiction. So $\phi$ is an automorphism that is not inner, and so $|\operatorname{Aut}(D_{12})| \ge 7$, so $|\operatorname{Aut}(D_{12})| = 12$.

From $r^{-j}=r^{j+1}$ I get $1=r^{2j+1}$ and $6 \,|\, (2j+1)$ which is a contradiction. (Sorry, I needed this line to see your conclusion.)

The case you dropped gives $sr^{j}s=r^{-j}=r^{j+1}$. (Sorry, I didn't see immediately immediately.)

Yes. That's ok. Couldn't we generalize this for $\phi(r)=r\; , \;\phi(s)=sr^l$ to answer the question which automorphisms are outer and which are inner instead of the counting argument?