Let’s take stock of what we know about group theory so far in the first series.

- We defined a group, which is a set endowed with a binary operation satisfying 3 properties.
- For each group, we considered subsets which could inherit the group structure. These were called subgroups.
- Each subgroup neatly partitioned the ambient group into (left/right) cosets, which gives rise to Lagrange’s theorem (among other things).
- If the subgroup is
*normal*, then the collection of cosets forms a group, called the*quotient group*. - Homomorphisms are functions between groups which respect the underlying operation. The kernel (set of elements mapping to the identity) is a normal subgroup, while the image is a subgroup. By the first isomorphism theorem, every homomorphism is effectively a composition of projection
*G*→*G*/*N*, composed with injective*G*/*N*→*H*.

These are basic properties of a group which seem rather natural to consider. What other interesting questions of groups can we pose?

Question 1: is the converse to Lagrange’s theorem true?

The theorem states that a subgroup *H* of *G* satisfies |*H*| | |*G*|. Conversely, if |*G*| = *n* has a factor *m*, one asks if *G* must have a subgroup of order *m*.

The answer is *no* but finding a counter-example is no easy affair: the smallest example I can think of is of order 60, and *m*=30. This follows from the two observations:

- has no normal subgroup except {
*e*} and itself; - any subgroup
*H*≤*G*of index 2 is normal (because if , then*G*is the disjoint union of*H*and*gH*, and also the disjoint union of*H*and*Hg*; thus*gH*=*Hg*).

Thus, our naive conjecture turned out to be false. But instead of giving up, let’s consider circumstances where this is true. It turns out, *if m is a prime-power which divides n = |G|, then G has a subgroup of order m*. This is part of Sylow’s theorems, which offers great insight into the structure of finite groups and their subgroups.

Question 2: how many groups, up to isomorphism, are there of ordernforn= 1, 2, … ?

When *n* is prime, we already know the answer: the group is cyclic. With a bit of work, one can prove later that if *n* = *p*^{2} (*p* prime), then the only possibilities are **Z**/*p*^{2} and (**Z**/*p*) × (**Z**/*p*). Things quickly get worse as we consider higher powers of prime.

An interesting phenomenon is that we get the largest number of groups for 2^{k}. To be specific, for *n* close to 2^{k}, there are far more groups of order 2^{k} than not. In fact, among all groups of order at most 2000, more than 99% of them have order 1024. And if we exclude these, more than 96% of the remaining have order 768.

Question 3: can we consider the structure of a group by “factoring” it into its constituents?

If *N* is a normal subgroup of *G*, then we obtain subgroup *N* and quotient group *G*/*N*. Conversely, one hopes to reconstruct *G* by looking at the smaller two constituents *N* and *G*/*N*. The question then becomes the following.

Question 4. GivenNandH, how do we construct all groups G containing N such that ?

The extension problem is theoretically solved, but actually computing all the extensions can be a real hassle. We hope to go through some explicit examples by constructing extensions of groups of order 2^{n}.

But even if the extension problem is solved, there remains the problem of classifying groups with no nontrivial subgroups.

Question 5. A group of order > 1 is said to besimpleif its only normal subgroups are {e} and itself.[ We exclude the trivial group since we need simple groups as building blocks for more complicated groups. In this regard, the trivial group is useless. Just like 1 is not a prime number. ]

If *G* is abelian, then *G* is simple if and only if it has prime order (Exercise: prove it, using the fact that all subgroups of abelian groups are automatically normal.) For non-abelian groups, things get much stickier. The classification of finite simple groups was a massive endeavour undertaken by numerous mathematicians, which lasted until the mid 1980s. It was only twenty years later that a gap was found and had to be filled in. Here’s a more detailed overview of the history.

Speaking of breaking a group into constituents, one naturally asks if the set of constituents is unique. Explicitly:

Question 6. Let G be a finite group. Write:,

where each and the quotient is simple. Is the set of unique up to isomorphism?

The answer is yes: this is the Jordan-Hölder theorem, which holds in far greater generality. Its proof is rather technical so we won’t be going through it.