Genady

I.e., show that A4 has a normal subgroup. I did it by brutal force and would like to know if there is a more elegant way. My solution: A4 consists of 4!/2=12 permutations: e = identity, 8 permutations of the kind (1 2 3)=(1 2)(2 3), (1 3 2)=(1 3)(3 2), etc., and 3 permutations with separated cycles: a = (1 2)(3 4) b = (1 3)(2 4) c = (1 4)(2 3) Because of the separation, the cycles in a, b, and c commute, and thus a2 = b2 = c2 = e. I've checked manually that ab = ba = c, ac = ca = b, and bc = cb = a. So, {e, a, b, c} is an abelian group. An abelian subgroup is normal. Thus, A4 is not simple. Let me know if any of the above need elaboration.

Show that Sn is isomorphic to a subgroup of An+2. I will demonstrate my idea for n=3. S3 has 3!=6 permutations, 3 odd: (1 2), (1 3), (2 3); and 3 even: identity, (1 2 3)=(1 2)(2 3), (1 3 2)=(1 3)(3 2). Let's consider them separately. Put elements 4 and 5 in A5 aside. Identify even permutations in S3 with permutations in A5 with the same cycles as in S3 while the elements 4 and 5 are fixed, e.g., (1 2)(2 3) in S3 ↔ (1 2)(2 3) in A5. Identify odd permutations in S3 with permutations in A5 with the same cycles as in S3 plus the cycle (4 5), e.g., (1 2) in S3 ↔ (1 2)(4 5) in A5. I think, it is obvious how to generalize it for any n, right?

Show that An for n≥3 is generated by 3-cycles, i.e., any element can be written as a product of 3-cycles. My solution is quite simple, but I wonder if I've missed something: Any element of an alternating group can be written as a product of even number of 2-cycles. Let's consider pairs of 2-cycles in such expression. They can be of two forms: (a,b)(b,c) and (a,b)(c,d). The first is immediately a 3-cycle: (a,b)(b,c)=(a,b,c). The second can be made into 3-cycles like this: (a,b)(c,d)=(a,b)(b,c)(b,c)(c,d)=(a,b,c)(b,c,d). So, each pair of 2-cycles can be converted to one to two 3-cycles.

The word "love" appeared there in a translation and by necessity an interpretation of something written in a dead language.

My point is that you pick an interpretation as you like. They are only words. Humans make a meaning.

Mistranslations and misinterpretations. as discussed here: “Love Your Neighbor as Yourself”—What It Really Means - The BAS Library

If I "ever wondered why ice cream has that perfectly smooth and creamy texture," I'd write the question in Google and get the answers.

This lines also relate only to the chosen people.

Only as far as "the chosen people" were concerned.

This is the AI that failed my tests, e.g., https://www.scienceforums.net/topic/135352-help-to-share-beauty-of-math-with-more-people/#findComment-1283477 https://www.scienceforums.net/topic/135377-are-tangent-points-allowed-in-ruler-and-compass-constructions/#findComment-1283623

\(a+b\)

Solved. The answer to the OP question is yes, because I've found a simple way to convert a "touching case" into a "crossing case."

And another AI fail. A meaningless "solution": ### Steps to Construct Point C (Midpoint of A and B) Using Only a Compass: 1. **Draw a circle centered at A with radius AB:** - Place the compass at point A and adjust its width to the distance AB. - Draw a circle centered at A with radius AB. 2. **Draw a circle centered at B with radius AB:** - Without changing the compass width, place the compass at point B. - Draw a circle centered at B with radius AB. 3. **Find the intersection points of the two circles:** - The two circles will intersect at two points. Let’s call these points P and Q. 4. **Draw a circle centered at P with radius PA:** - Place the compass at point P and adjust its width to the distance PA (which is equal to AB). - Draw a circle centered at P with radius PA. 5. **Draw a circle centered at Q with radius QA:** - Similarly, place the compass at point Q and adjust its width to the distance QA (which is also equal to AB). - Draw a circle centered at Q with radius QA. 6. **Find the intersection points of the new circles:** - The circles centered at P and Q will intersect at two points: one of these points is A, and the other is the midpoint C. 7. **Identify the midpoint C:** - The intersection point that is not A is the midpoint C between A and B. ---

I've found that the last problem can be solved also by using intersections of circles, i.e., circles crossing rather than touching at points. It can be done by following constructions in the proof of Mohr–Mascheroni theorem - Wikipedia. Although, these constructions are very long and convoluted.

Saying that the emergence of life was "guided" puts them squarely into the "divine intervention" camp.

Good, but this is not the problem that led to the question in the thread. Here is a solution of this problem: Now, the problem in question is this: Given points A and B construct point C in the middle between them using only compass. I've solved this problem, but my solution relies on being able to mark a point where two circles touch rather than cross each other. I don't know if this is legitimate. That was why I've asked.

I should just formulate the original problem, and all will become clear: Given two points, A and B, construct a point C, using only a compass, such that B is a midpoint between A and C.

Yes, but as I've clarified in the post before yours, I don't have a centerline, or any line, because I don't have a ruler. If I had ruler, I would use it only as a straightedge, i.e., in the classical meaning. The circle is drawn, and its center is known, so it is not a problem to take its radius with compass. Then, the diameter can be constructed by doubling the radius, which is doable with compass only, i.e., no ruler is needed for this.

Even more specifically... With ruler and compass I could construct the point where the circles touch as a point where the line connecting their centers cross them. However, what I deal with is a construction using compass only, no ruler. I don't know if constructing that point by crossing only some circles is possible.

In classical constructions with ruler and compass, a construction proceeds from point to point where the points are intersections of lines and circles, i.e., where the lines and the circles cross each other. Are the points where the lines and the circles touch each other rather than cross, allowed as well? More specifically, if I have constructed two circles and I know that they touch at some point, can I proceed with the construction using this point?

Fair enough.

Do you think it is possible that they ask a question in a math test about something that has not been defined?

It takes 1 hour.

Not only its answer is wrong, but also its first statement: because a semicircle is rather one-dimensional. Incidentally, I've just read that this AI, DeepSeek, (Tech leaders respond to the rapid rise of DeepSeek | VentureBeat). So, they are all that bad. P.S. It seems that you think I brought the AI answer as a correct one. Clearly, there is a miscommunication, because my intent was exactly opposite. P.P.S. I make mistakes in English from time to time, but I do not translate from Russian.

I don't know why it would matter, unless they were taught something wrong. I also don't see anything tricky about the question. They are given a figure, which consists of a semicircle, a line, and an area, and they need to answer the question about the semicircle.