## Curiosity Number One

The additive group of polynomials with integer coefficients is isomorphic to the multiplicative group of positive rational numbers: (ℤ[x], +) ≅ (ℚ_{+}, ⋅).

Any element of ℤ[x] can be written like a_{0} + a_{1}x + a_{2}x^{2} + … + a_{n}x^{n}, where all coefficients a_{0}, a_{1}, …, a_{n} are integers. Consider the following mapping: φ: (a_{0} + a_{1}x + a_{2}x^{2} + … + a_{n}x^{n}) ↦ (p_{0}^{a0}⋅p_{1}^{a1}⋅…⋅p_{n}^{an}), where p_{k} refers to the kth prime number. Now, this is a homomorphism from ℤ[x] to ℚ_{+} because φ(p(x) + q(x)) = φ(p(x))⋅φ(q(x)). You can also easily show that it is onto and one-to-one, using the uniqueness of prime factorization. Therefore φ is an isomorphism!

I think this is a good example to test the intuitive notion that once one “knows” a group, one also knows all groups that it is isomorphic to. It’s often useful to think of isomorphic groups as being like one book written in two different languages, with the isomorphism being the translation dictionary. But the non-obviousness of the isomorphism here makes it feel like the two groups in fact have a considerably different internal structure.

## Curiosity Number Two

Your friend comes up to you and tells you, “I’m thinking of some finite group. All I’ll tell you is that it has at least two order-2 elements. I’ll give you $20 if you can tell me what the group that’s generated by these two elements is (up to isomorphism)!“

What do you think your chances are of getting this $20 reward? It seems intuitively like your chances are probably pretty dismal… there are probably hundreds or maybe even an infinity of groups that match her description.

But it turns out that you can say exactly what group your friend is thinking of, just from this information!

Call the group G. G is finite and has two order-2 elements, which we’ll call a and b. So we want to find out what <a,b> is. Define r = ab and n = |r|. Then ara = a(ab)a = (aa)ba = ba = (ab)^{-1} = r^{-1} = r^{n-1}. So ara = r^{n-1}, or in other words ar = r^{n-1}a. Also, since b = ar, we can write the group we’re looking for either as <a,b> or as <a,r> (these two are equal).

Thus the group we’re looking for can be described as follows:

<a, r | a^{2} = r^{n} = e, ar = r^{n-1}a>.

Look familiar? This is just the dihedral group of size 2n; it is the group of symmetries of a regular n-gon, where a is a flip and r is a rotation!

So if G is a finite group with at least two order-2 elements a and b, then <a, b> ≅ Dih_{|ab|}.

This serves to explain the ubiquity and importance of the dihedral groups! Any time a group contains more than one order-two element, it must have a dihedral group of some order as a subgroup.

## Curiosity Number Three

If K ◃ H ◃ G, then H/K ◃ G/K and (G/K) / (H/K) ≅ G/H.

(G/K) / (H/K) is a pretty wild object; it is a group of cosets, whose representatives are *themselves* cosets. But this theorem allows us to treat it as a much less exotic object, an ordinary quotient group with representatives as elements from G.

I don’t have much else to say about this one, besides that I love how the operation of forming a quotient group behaves so much like ordinary division!