Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need help on question 1 I am preparing for my test:

1) Find a homomorphism $\alpha: A_4 \to\mathbb Z_6$ such that $\ker (\alpha) = K$ where $K$ is the normal subgroup $K= \{(1), (12)(34), (13)(24), (14)(23)\}$.

I am so confused on this I know the order of $A_4$ is $12$ and I listed all of them down and then found their orders but I am confused in how to obtain the answer.

share|improve this question

2 Answers 2

To get a feel for what this homomorphism should be, first realise $A_4$ as the rotational symmetry group of the tetrahedra. For this, you label each vertex of the tetrahedra with a number from 1 to 4. Then a rotational symmetry of the tetrahedra corresponds to a permutation of its vertices, hence a permutation of $\{1,2,3,4\}$. You can check that doing this gives you an isomorphism between $A_4$ and the group of rotational symmetries of the tetrahedra.

Now, the subgroup $K$ you give is a particular subgroup of the rotations of the tetrahedra. It corresponds to those rotations where you fix two middle points of opposed edges and you flip the whole thing by an angle of $\pi$. The only other rotations are the ones where you fix a vertex and rotate by an angle of $2 \pi /3$ with respect to the axis joining this vertex and the center of the opposed face.

Then, by the first isomorphism theorem, the image of $A_4$ in $\mathbb{Z}_6$ by our homomorphism will be $\mathbb{Z}_3$ and we need to view the remaining rotations (those of angle $2 \pi /3$, fixing a point) modulo $K$. One way to do this is to identify every counter-clockwise rotations (call these positive) together and do the same thing with clockwise (call these negative) rotations. Note that this orientation is given by looking at the tetrahedra from the point of view of the fixed point, for example, draw your tetrahedra as sitting on a table, with the face touching the table having a vertex going away from you. Now label the upper vertex 1, the left-most vertex touching the table 2, the right-most 3 and the one in the back 4. Then the rotation fixing 1, sending 2 to 3, 3 to 4 and 4 to 2 is a counter-clockwise, or positive rotation.

The homomorphism in question will be sending the positive rotations to $2$ in $\mathbb{Z}_6$, the negative rotations to $4$, and elements of $K$ to $0$. We need to check that this is indeed a homomorphism. To see this, note that composing positive (resp. negative) rotations by elements of $K$ gives positive (resp. negative) rotations. Also, $K$ acts on positive/negative rotations in a transitive way (meaning that from a positive rotation you can get any other positive rotation by composing with an element of $K$). Indeed, you can check these by multiplying the corresponding permutations in cyclic notation. This means that for a positive rotation, you can always pick the fixed point of your choice. With this out of the way, it's easy to see that composing two positive rotations gives a negative rotation, that two negative gives a positive, and that a positive composed with a negative gives an element of $K$. Here's how:

Let $p, p'$ be two positive rotations, then there exists $k \in K$ such that $p \circ p' = p \circ p \circ k$. But the composition of a positive rotation with itself is certainly negative. The argument is the same with negative rotations. For the last assertion, let $p$ be a positive rotation and $n$ be a negative rotation. Then $p \circ n = p \circ p^{-1} \circ k = k$ for some $k \in K$.

Explicitly, the positive rotations are $\{(143),(234),(124),(132)\}$ and the negative rotations are $\{(243),(134),(142),(123)\}$. So an answer to your question is $$\{(143),(234),(124),(132)\} \mapsto 2 \in \mathbb{Z}_6$$ $$\{(243),(134),(142),(123)\} \mapsto 4 \in \mathbb{Z}_6$$ $$K \mapsto 0 \in \mathbb{Z}_6$$.

Note that sending the negative rotations to 2 and the positive rotations to 4 would also give you an answer.

share|improve this answer

$K$ is a normal subgroup and the quotient $A_4/K$ is cyclic of order 3, $C_3$, say generated by $x$. There is a unique homomorphism $\phi: C_3\rightarrow Z_6$, determined by $x\rightarrow 2$. Let $\pi$ denote the natural homomorphism $A_4\rightarrow A_4/K$. The homomorphism $\phi\pi: A_4\rightarrow Z_2$ has kernel $K$ and image $\{0,2,4\}$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.