# Thread: some more problems about groups

1. ## some more problems about groups

1. Define the mobius group, and describe how it acts on C U {infinity} , U is union.

Show that the subgroup of the mobius group consisting of transformations which fix 0 and infinity is isomorphic to C\{0}

Now show that the subgroup of the mobius group consisting of transformations which fix 0 and 1 is also isomorphic to C\{0}

2. let G be a dihedral group of order 12
i) list all the subgroups of G of order 2. Which of them are normal
ii) list all the remaining proper subgroups of G
iii) for each proper normal subgroup N of G , describe the quotient group G/N
iv) show that G is not isomorphic to the alternating group A4

3. Show that if a group G contains a normal subgroup of order 3, and a normal subgroup of order 5, then G contains an element of order 15.

Give an example of a group of order 10 with no element of order 10.

4. suppose that G is the group of rotational symmetries of a cube C. Two regular tetrahedra T and T' are inscribe in C, each using half the vertices of C. what is the order of the stabilizer in G of T?

5. Suppose that G is a finite group of rotations in R2 about the origin. Is G cyclic? justify your answer.

6. G X H = { ( g,h): g belongs to G, h belongs to H}
show how to make G X H into a group in such a way that G X H contains subgroups isomorphic to G and H.

2. Originally Posted by szpengchao
1. Define the mobius group, and describe how it acts on C U {infinity} , U is union.
You can define the action of a Mobius transformation on a point in (the Riemann sphere) $\mathbb{C}_{\infty}$ to be the image of that point under the Mobius transformation.

Show that the subgroup of the mobius group consisting of transformations which fix 0 and infinity is isomorphic to C\{0}
Consider $f(z) = (az+b)/(cz+d)$ where $ad-bc\not = 0$. Since $f(\infty) = \infty$ it means $c=0$ and we simply have $f(z) = rz+s$. Since $f(0)=0$ it means $f(z)=rz$ where $r\not = 0$. Thus, this set is the set of all non-zero dilations $G=\{ rz : r\in \mathbb{C}^{\text{x}} \}$ and this isomorphic to the complex group under multiplication under the isomorphism $\phi: G\mapsto \mathbb{C}^{\text{x}}$ as $\phi(rz) = r$.

Now show that the subgroup of the mobius group consisting of transformations which fix 0 and 1 is also isomorphic to C\{0}
It is similar to what was done above.

3. Originally Posted by szpengchao
2. let G be a dihedral group of order 12
i) list all the subgroups of G of order 2. Which of them are normal
ii) list all the remaining proper subgroups of G
iii) for each proper normal subgroup N of G , describe the quotient group G/N
iv) show that G is not isomorphic to the alternating group A4
Note $\text{D}_6 = \{ 1,x,...,x^5,y,xy,...,x^5y\}$ with $yx=x^5y$ and $x^6=y^2 = 1$. For part (i) find the elements which have order 2. To do that suppose an element has form $x^a$ then we want $(x^a)^2 = 1\implies x^{2a} = 1$ so $6|2a \implies 3|a$. Thus, $a=3$ which means $x^3$ is the only element with order $2$. Similarly, now look which element of form $x^ay$ have order $2$ thus we want $(x^ay) = 1\implies x^ayx^ay = 1\implies x^ax^{-a}y^2 = 1\implies 1=1$. Thus, for any $a$ we have order two. This means the order two subgroups are $\{1,x^3\}, \{1,y\},\{1,xy\},...,\{1,x^5y\}$. Similarly, order three subgroups are found in the same manner. We want $(x^a)^3 = 1\implies a=2$ which gives us $\{1,x^2,x^4\}$. While $(x^ay)^3 = (x^ay)^2(x^ay) = x^a y\not = 1$. Thus, $\{1,x^2,x^4\}$ is the only order three subgroup. And then you need to go to order 6 subgroup, the obvious one is $\{1,x,...,x^5\}$ but it would take work to find the other ones (if there are other ones). Once you have that you can easily do the second and thrid part. Look which one are normal and form quotients. And compare with $A_4$ subgroup structure which is different, such as it has no element of order 6.

4. Originally Posted by szpengchao
3. Show that if a group G contains a normal subgroup of order 3, and a normal subgroup of order 5, then G contains an element of order 15.
First note that $H\cap K = \{ 1\}$. Let $x\in H$ and $x\in K$ then the order of $x$ divides $|H|=5$ and $|K|=3$. This means $x=1$ because $\gcd(3,5)=1$, thus $H\cap K = \{1\}$. The next property is kind of supprising, $hk=kh$ for all $h\in H$ and $k\in K$. Note $hkh^{-1}k^{-1}\in H$ because $h\in H,kh^{-1}k^{-1}\in H$ and $hkh^{-1} \in K, k^{-1}\in K$ because they are normal. Since the intersection is trivial it means $hkh^{-}k^{-1} = 1\implies hk=kh$. Now let $x$ be generator of $H$ and $y$ be generator of $K$ and consider $xy$. Note $(xy)^k = x^k y^k$ because $xy=yx$. This means the order of $xy$ is $\mbox{lcm}(3,5)=15$.

5. Originally Posted by szpengchao
6. G X H = { ( g,h): g belongs to G, h belongs to H}
show how to make G X H into a group in such a way that G X H contains subgroups isomorphic to G and H.
Just define $(g_1,h_1)(g_2,h_2) = (g_1h_1,g_2h_2)$. Then $G\times H$ is a group. And $A = \{ (g,e)|g\in G\}$ is subgroup isomorphic with $G$. Similarly with $H$.

6. Originally Posted by szpengchao
5. Suppose that G is a finite group of rotations in R2 about the origin. Is G cyclic? justify your answer.
.
It seems that $G$ has to be cyclic. Instead of thinking of these as rotations of $\mathbb{R}^2$ we will think of them as rotations in the complex plane $\mathbb{C}$. Thus, we can think of these rotations are $e^{i\theta}$ where $-\pi < \theta \leq \pi$, where the positive sign indicates counterclockwise rotation and negative sign indicates clockwise rotation. Now if $e^{i\theta} \in G$ it must have finite order, suppose $n$ is the order than $e^{in\theta}$ has to be the identity, i.e. $in\theta = 2k\pi i\implies \theta = \frac{2\pi k}{n}$. Therefore, this shows that these rotations must be multiples of $2\pi$ divided by an integer. Now consider $e^{2\pi ik_1/n_1},...,e^{2\pi ik_j/n_j}$ be the group of rotations (excluding the identity). Define $m=n_1\cdot ... \cdot n_j$. And define the group of rotations generated by $e^{2\pi i/m}$. Then this group is cyclic and $G$ is a subgroup of this group, since a subgroup of a cyclic group is cyclic it means $G$ must be cyclic.

(But in general finite symettry groups in $\mathbb{C}$ do not have to be cyclic).