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Could someone give me a rigorous proof that the group of rotations each element of which is a composition of rotations around the altitudes of a tetrahedron that transform the tetrahedron into itself is isomorphic to the alternating group $A_4$.

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Number the vertices. See what each rotation does to the vertices. Write them as permutations. – Arturo Magidin Jun 6 '12 at 20:59
So far, I've been trying to understand what does group of rotations mean. – Adam Jun 6 '12 at 21:02
Take a tetrahedron and put it on a table. Pick it up and put it down again so it is not necessarily the same way up, but occupies the same position in space. You will find that the same effect could have been achieved by a rotation - there are two kinds of axis: one kind through a vertex and the centre of the opposite face, and another kind through the middle of two opposite edges. – Mark Bennet Jun 6 '12 at 21:11
up vote 2 down vote accepted

Jykri's solution is excellent, but it may be useful to have another way of understanding why the symmetry group isn't all of $S_4$ (since it may not be immediately obvious that the group is generated by its three-cycles - that is, that every element of the group can be written as a product of three-cycles). Choose one of the vertices, $v_0$ say, and consider the triple product of the vectors from $v_0$ to the three other vertices $v_1, v_2, v_3$: $((v_1-v_0)\times(v_2-v_0))\cdot(v_3-v_0)$. This is the determinant of the matrix whose rows are the three vectors $v_i-v_0$; it represents the signed volume of the tetrahedron. As such, its sign has to be preserved by any rotation of the tetrahedron (this just says that rotations preserve the orientation of space), but a little calculation will show that swapping two vertices changes the sign of the triple product. This means that the two-cycles aren't in the rotation group, and so the group cab't be all of $S_4$ (since we know some elements of $S_4$ that aren't in it). Jykri's answer shows that the group is at least as large as $A_4$, and $A_4$ is a so-called maximal subgroup of $S_4$; there are no other groups 'in between' $A_4$ and $S_4$. Since the rotation group is at least $A_4$ and it can't be anything larger, it must be $A_4$ exactly.

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I think I got the idea. How can I formally prove that it is isomorphic to A4 though? – Adam Jun 6 '12 at 22:47
@AdamAndersson What do you consider a 'formal' proof, and what do you think is missing from the outlines you've already been given? Do you understand why it must be some subgroup of $S_4$? If so, then you're already almost all the way there; Jykri's argument shows why it must be 'at least' $A_4$, and mine shows why it's not $S_4$. – Steven Stadnicki Jun 6 '12 at 22:53

The simple approach may be the most illuminating:

Draw a picture.

Label the vertices (there are four of them, so it may be a good idea to use numerals 1, 2,3, 4).

Check the effects of the 120 degree rotations on the labels of the vertices (when done correctly, you should get 3-cycles).

Start combining those rotations until you get everything (this is also a useful exercise for the purposes of becoming fluent in the art of combining permutations).

If you don't want to do all those combinations, then you can replace the last step with the following piece of reasoning. We extract two facts from the above: 1) that the group of rotations can be viewed as a subgroup of $S_4$, because any rotation is fully determined by its effect on the 4 vertices, 2) the generators are all even permutations (3-cycles), so the group in question is actually a subgroup of $A_4$.

The group $A_4$ has 12 elements, so it suffices to show that we can get at least 12 different permutations of vertices by rotating the tetrahedron (there can't be more than twelve). First let's check that we can rotate any one of the four corners to the top vertex (imagine the tetrahedron lying on one side on a table, so that there is a unique top vertex). To do this we simply pick one of the vertices that is neither the top vertex nor the other vertex that we want to bring to the top, and then grab that vertex and rotate the thing by 120 degrees in the appropriate direction withour releasing our grip. Ok, so there are 4 choices for the vertex at the top. But we can then grab that top vertex and rotate the "bottom" face (the one touching the table). So we can bring the remaining three vertices into three different positions while keeping the top fixed.

We have shown that we can permute the vertices of a tetrahedron into $4\cdot3=12$ different positions, so by our earlier observations these permutations must form the group $A_4$.

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I will answer the question implied by your comment, "what is a group of rotations".

The point of group theory is to study symmetries of mathematical objects, including geometric objects. For example, a tetrahedron.

If you have a tetrahedron sitting on the table so that one vertex is pointing straight up, you can rotate that tetrahedron 120° around its vertical axis and it will look just the same. This motion is called a symmetry of the tetrahedron. Any motion of the tetrahedron that leaves it looking exactly the same is a "symmetry".

It should be clear that performing two symmetries in a row is itself a symmetry. The first symmetry leaves the tetrahedron looking the same, and so does the second, and therefore the combined motion leaves them looking the same. This "composition" combines two symmetries into one. For example, turning the tetrahedron clockwise once, and then clockwise again, is the same as turning it counterclockwise once. If $p$ and $q$ are symmetries, we might write $p\cdot q$ for the motion, also a symmetry, that first does $p$ and then does $q$.

It's perhaps not perfectly clear, but it is true, that for any symmetries $p$, $q$, and $r$, $(p\cdot q)\cdot r = p\cdot(q\cdot r)$. Thus the composition of symmetries of a tetrahedron is an associative operator on the symmetries. (This is actually a special case of the fact that composition of functions is always associative.)

It should be clear that there's a "trivial" symmetry that makes no motion at all. This non-motion certainly leaves the tetrahedron looking exactly the same, so it is indeed a symmetry. If $e$ is the do-nothing symmetry, then $e\cdot p = p$ and $p\cdot e = p$ for every symmetry $p$. That is, the "do nothing" symmetry behaves like an identity element with respect to composition of motions.

It should be clear that for each symmetry, there is an opposite motion, also a symmetry, that undoes its effect. If $q$ is the "opposite" of $p$, then $p\cdot q = e$ and $q\cdot p = e$. That is, doing the "opposite" motion behaves like a group inverse with respect to composition of motions.

So the symmetries of a tetrahedron form a group! And if this is a surprise, it is because your course has put the concepts backwards, because the reason that the axioms of a group are what they are, and not different axioms, is because these axioms were designed exactly to model these properties of the idea of symmetry.

The group is called "the group of rotations of a regular tetrahedron", because we considered only rotations of the tetrahedron. There is a larger group of symmetries that we could ascribe to a tetrahedron, by allowing mirror reflections of the tetrahedron as well as rotations. This might be called "the group of rotations and reflections of a regular tetrahedron".

If one has studied abstract groups, as you have, one might ask which group this is. One can count up the symmetries of the tetrahedron. There turn out to be 12, because in picking up the tetrahedron and putting it down again in the same place you get 4 choices about which face to put down on the table, and then 3 choices about which face to turn to face frontward. But there is more than one group with 12 elements. There are actually 5 such groups—there are 5 ways that an object can have exactly 12 symmetries.

So your job is to find out in which of these 5 ways the tetrahedron is symmetric. Is is symmetric in the way that a regular hexagon is symmetric, which we call $D_{12}$? Or is it symmetric in the (different) way that a 12-toothed saw blade is symmetric, which we call $Z_{12}$? Or is it, as the question claims, symmetric in the (different again) way that we call $A_4$?

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