# Thread: automorphism and dihedral group

1. ## automorphism and dihedral group

1) Descibe explicitly all homomorphisms
h: C_6 ---> Aut(C_12)

C_6 ={1,x,....,x^5} x^6 = 1

We know that Aut(C_12) = {q_1,q_5,q_7,q_11} =~ C_2*C_2

where * = "cross" is the direct product

Then, h: C_6 ---> C_2*C_2

So, I found that the possibilities for h(x) are:

h_1(x) = 1
h_2(x) = q_5
h_3(x) = q_7
h_4(x) = q_11

2)Describe explicitly all homomorphisms
h: C_5 ---> Aut(C_11)

Since 11 is prime number, so Aut(C_11) =~ C_10

I worked out that there are 5 homomorphism for C_5 ---> Aut(C_11). Is that true?

3)This question I dont know how to do it.

Show that Aut(C_2*C_2) =~ D_6 ( from cyclic group to dihedral group)

2. Originally Posted by knguyen2005

1) Descibe explicitly all homomorphisms
h: C_6 ---> Aut(C_12)

C_6 ={1,x,....,x^5} x^6 = 1

We know that Aut(C_12) = {q_1,q_5,q_7,q_11} =~ C_2*C_2

where * = "cross" is the direct product

Then, h: C_6 ---> C_2*C_2

So, I found that the possibilities for h(x) are:

h_1(x) = 1
h_2(x) = q_5
h_3(x) = q_7
h_4(x) = q_11

i don't know what you mean by $q_5, q_7, q_{11},$ but yes, there are exactly 4 homomorphisms, because a homomorphism can send $x,$ the generator of $C_6,$ to any of 4 elements of $C_2 \times C_2.$

2)Describe explicitly all homomorphisms
h: C_5 ---> Aut(C_11)

Since 11 is prime number, so Aut(C_11) =~ C_10

I worked out that there are 5 homomorphism for C_5 ---> Aut(C_11). Is that true?
correct!

3)This question I dont know how to do it.

Show that Aut(C_2*C_2) =~ D_6 ( from cyclic group to dihedral group)

let $G=C_2 \times C_2=\{1,x,y,xy\},$ where $x^2=y^2=1, \ xy=yx.$ let $f \in \text{Aut}(G).$ then $f(1)=1, \ f(xy)=f(x)f(y).$ now $f(x)$ can be defined to be any element of the set $G-\{1\}.$

since $f$ is one-to-one, $f(y)$ can be defined to be any element of the set $G-\{1,f(x)\}.$ so there are 3 possibilities for $f(x)$ and 2 possibilities for $f(y).$ therefore $|\text{Aut}(G)|=6.$

define the automorphisms $f,g$ by: $f(1)=1, \ f(x)=y, \ f(y)=x, \ f(xy)=xy,$ and $g(1)=1, \ g(x)=xy, \ g(y)=x, \ g(xy)=y.$ then $f \neq fg$ and $o(f)=o(fg)=2.$ hence $\text{Aut}(G)$

is not cyclic. now use this fact that if $p$ is a prime number, then any group of order $2p$ is either cyclic or it's isomorphisc to $D_{2p}.$

alternatively, see that $o(g)=3$ and $D_6 \simeq H=\{\text{id}_G, f, g, g^2, fg, fg^2 \} \subseteq \text{Aut}(G).$ [you need to check that all elements of $H$ are distinct.] but we have $|\text{Aut}(G)|=6.$ so: $\text{Aut}(G)=H.$

3. Question 1, cuz I was in rush, so I just used the short term

thus, Aut(C_12) = {q_1,q_5,q_7,q_11} =~ C_2*C_2
= { 1, a, b, ab} where a,b are generators of C_12
and a^2 = b^2 = 1
ab = ba
So, the homomorphism are:
h_1(x) = 1
h_2(x) = a
h_3(x) = b
h_4(x) = ab

Does that make sense now?

Question 3. Firstly, I'd like to thank NonCommAlg , the work is very good but there are some points I dont understand and I hope to get the explanation.

a) let where
I was taught that Aut (C_n) is Abelian but in general Aut(G) is non-abelian , e.g. Aut(C2*C2) =~D_6 . So, if you let G = C2*C2 then it is abelian group because xy=yx (commutative) . I'm confused

b) can be defined to be any element of the set so there are 3 possibilities for and 2 possibilities for therefore
When you said How do you work out the order of this group is 6 ( Id is also included, right).

c) then and hence is not cyclic
How do you know Aut(G) is not cyclic group and why do you have to use 2 automorphisms f and g. Is that because you want to find the generators of D_6.

d) why is ? and How do you come up with this?

I'm sorry because I got too many questions, I just want to make it clear.

Thank you so much

KN

4. ok, i'll answer some of your questions. you need to find the answer to the rest yourself by reading my solution more carefully!

Originally Posted by knguyen2005
Question 1, cuz I was in rush, so I just used the short term

thus, Aut(C_12) = {q_1,q_5,q_7,q_11} =~ C_2*C_2
= { 1, a, b, ab} where a,b are generators of C_12
and a^2 = b^2 = 1
ab = ba
So, the homomorphism are:
h_1(x) = 1
h_2(x) = a
h_3(x) = b
h_4(x) = ab

Does that make sense now?
yes, it's correct! just note that $a,b$ are the generators of $C_2 \times C_2$ not $C_{12}.$

a) let where
I was taught that Aut (C_n) is Abelian but in general Aut(G) is non-abelian , e.g. Aut(C2*C2) =~D_6 . So, if you let G = C2*C2 then it is abelian group because xy=yx (commutative) . I'm confused
the automorphism group of an abelian group might be abelian or non-abelian. the purpose of this exercise is to show you that there are abelian groups G such that Aut(G) is non-abelian.

b) can be defined to be any element of the set so there are 3 possibilities for and 2 possibilities for therefore
When you said How do you work out the order of this group is 6 ( Id is also included, right).
an automorphism $f$ is completely determined if we know $f(x)$ and $f(y),$ because $f(1)$ has to be 1 and if we know $f(x),f(y),$ then we'll also know $f(xy)=f(x)f(y).$ now $f(x)$ has 3 possible values

and $f(y)$ has two possible values. in order to define an automorphism, we have to choose $f(x)$ from one of those 3 possibilities and $f(y)$ from those two possibile values. obviously that can happen

in 6 ways.

c) then and hence is not cyclic

How do you know Aut(G) is not cyclic group and why do you have to use 2 automorphisms f and g. Is that because you want to find the generators of D_6.
because a cyclic group of order 6 has exactly one element of order 2.