## Cosets and Lagrange’s Theorem

[ This post approximately corresponds to chapter IV from the old group theory blog. ]

The main theorem in this post is Lagrange’s theorem: if H ≤ G is a subgroup then |H| divides |G|.

But first, let’s consider cosets. Now, when H is a subgroup of G, it turns out G can be “sliced up” into subsets of equal sizes, one of which is H itself. To put it explicitly, let’s define a (leftcoset to be a set of the form:

$gH = \{ gh : h\in H\}$, where $g\in G$.

Then it turns out two cosets are either equal or disjoint:

Proposition. Let $x, y \in G$ be any elements.

• If $x^{-1}y \in H$, then $xH = yH$.
• If $x^{-1}y \not\in H$, then $xH \cap yH = \emptyset$.

The proof is quite easy, but it’s here if you want it. Diagrammatically this gives:

Furthermore, we have |gH| = |H| for any g since the left-multiplication map which takes x to gx is a bijection between H and gH. Immediately, we obtain:

Lagrange’s Theorem. If H is a subgroup of finite group G, then |H| divides |G|. Also, |G|/|H| = [G:H], the number of (left) cosets of H in G. We call this the index of H in G.

In particular, if g is an element of G, then $\left$ is a subgroup of order o(g), so o(g) must divide |G|. Thus, $g^{|G|} = e$.

The theorem is rather surprising at first glance, since in the case of subset of a set, one can only say whether an element is in or not in the subset. In the case of subgroups, we get a nice partitioning.

Note: the following picture may help. If $G=\mathbf{R}^3$ and H is a plane in G passing through the origin, then each coset is just a translate of the plane; so $\mathbf{R}^3$ is the union of disjoint planes parallel to H.

## Examples

Here’re the examples of subgroups from the last post. Let’s check that they satisfy Lagrange’s theorem.

1. For the subgroup mZZ, the index [ZmZ] = m if m>0. In the case m=0, the index is infinite.
2. For Q*2 < Q>0 < Q*, the index [Q* : Q>0] = 2 since we can just pick {+1, -1} as coset representatives, i.e. +1 and -1 belong to different cosets and the two form a disjoint union. On the other hand [Q>0 : Q*2] is infinite: can you find a good set of coset representatives? Answer (highlight to read): the set of primes.
3. For the subgroup k·(Z/m) < Z/m, where k|m, the index equals k since the subgroup has order m/k.
4. For (Z/21)*2 < (Z/21)*, the RHS group has elements {1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20} while LHS group has elements {1, 4, 16}. Thus, the index = 12/3 = 4, with coset representatives {1, 2, 5, 10}.

Now for the non-abelian cases.

1. Since |Sm| = m! and |Sn| = n!, the embedding $S_m \to S_n$ for m ≤ n has index $P^n_{n-m} = n!/m!$.
2. Since |V| = 4 and |A4| = 4!/2 = 12, the index = 3. For coset representatives, we can pick  e, (1,2,3) and (1,3,2).
3. For A< Sn, the index = n!/(n!/2) = 2. Any odd permutation is a coset representative.

One particularly useful consequence is the following theorem in classical number theory.

Euler’s Theorem. Let a be coprime to m, where m>0. Then $a^{\phi(m)} \equiv 1 \pmod m$, where $\phi(m)$ denotes the Euler-totient function.

In particular, if p is prime and a is not a multiple of p, then $a^{p-1} \equiv 1 \pmod p$.

Indeed, the group (Z/m)* has order φ(m) so every element a in the group must satisfy $a^{\phi(m)}\equiv 1 \pmod m$. Thus, going into an abstract theory can sometimes cast a new light on classical results.

Another immediate result is the following.

Any group of prime order must be cyclic.

Indeed, let G be of prime order p and g be an element which is not the identity. By Lagrange’s theorem, o(g) divides p so it equals 1 or p. Since g ≠ e, it can’t be 1. So the order of g is exactly p and we must have $G = \left$ since both have order p.

Thus, groups of orders 1, 2, 3, 5 are all cyclic.

Exercise : prove that if G has order 4, then G is isomorphic to either Z/4 or (Z/2) × (Z/2).

Answer (highlight to read): if x is an non-identity element, then its order divides 4 so must be 2 or 4. If it’s 4, G is cyclic. Otherwise, x*x = e. Pick another element y other than e and x. Again, y*y = e. Now xy is different from ex, y, so G = {exyxy}. Same goes for yx, so xyyx. QED.

More generally, any group of order p2 (p prime) is either Z/p2 or (Z/p) × (Z/p) but it’s hard to prove it with our current set of tools. Inspired by this, consider a rather ambitious goal: can we classify all finite groups of a certain order? Although this question is too general and immense, it’s nice to keep it in mind when we look at later concepts, in particular the Sylow theorems.

Next, it’s possible to define right cosets: these are sets of the form:

$Hg = \{hg : h\in H\}$

for a subgroup H ≤ G. Again, any two right cosets are either disjoint or identical so we could have proven Lagrange’s theorem with right cosets too. The left and right cosets don’t coincide in general. For example, consider H = {e, (1,2)} < S3. If we use the coset representative g = (1,3), then:

gH = {(1,3), (1,2,3)}   and   Hg = {(1,3), (1,3,2)}.

When left and right cosets do coincide, something interesting happens, but we’ll leave this to the next post.

Finally, we mention briefly about double cosets: if HK ≤ G are both subgroups, then we define $HgK = \{hgk : h\in H, k\in K\}$. Again, two cosets are either identical or disjoint, but now their sizes can vary. This is not really an important topic, but interested readers can read about it here.

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