## MaBloWriMo 26: Left cosets

Let $G$ be a group and $H$ a subgroup of $G$. Then for each element $a \in G$ we can define a left coset of $H$ by $aH = \{ ah \mid h \in H \}$.

That is, $aH$ is the set we get by combining $a$ (on the left) with every element of $H$. For example, given the subgroup $\{0,4\} \leq \mathbb{Z}_8$ (this was the other subgroup of $\mathbb{Z}_8$—did you find it?), the left coset corresponding to $1 \in \mathbb{Z}_8$ is $1 +_8 \{0,4\} = \{1 +_8 0, 1 +_8 4\} = \{1,5\}$. A few observations:

• The cosets corresponding to different elements of $G$ might be the same. For example, the left coset of $\{0,4\}$ corresponding to $5 \in \mathbb{Z}_8$ is $\{5 +_8 0, 5 +_8 4\} = \{5, 1\} = \{1, 5\}$, just like the coset for $1$.
• Can you find the other possible (left) cosets of $\{0,4\}$ in $\mathbb{Z}_8$? What do you notice?
• As you may guess, there are also things called right cosets, denoted $Ha$, where we combine with an element on the right. $\mathbb{Z}_8$ is not such a good example anymore, since in $\mathbb{Z}_8$ the binary operation is commutative, that is, $a +_8 b = b +_8 a$, so left and right cosets are the same thing. In general, though, the binary operation of a group does not have to be commutative.
• There is nothing special about left cosets as opposed to right cosets. In our proof of Lagrange’s Theorem we will use left cosets, but we could equally well replace all the left cosets by right cosets (and flip a few other things around) to get a different but equally valid proof.
• As an interesting aside, when the left and right cosets of a subgroup coincide, we say that the subgroup is normal. (Hence every subgroup of a group with a commutative binary operation is normal; but this can also happen even when the binary operation is not commutative.) It turns out that these normal subgroups are very important. Normal subgroups of a group are kind of like the divisors of an integer; you can “divide” a group by one of its normal subgroups to get a “quotient group”. And yes, there are special groups called simple groups which don’t have any normal subgroups, and are kind of like prime numbers—there is a suitable sense in which every finite group can be uniquely decomposed into a “product” of simple groups, just like integers can be uniquely decomposed into a product of prime factors. But this is getting way off on a tangent! (I told you this proof would hint at some very cool, deeper group theory.)

Just one more observation for today. For any $a \in G$, consider the function $f_a(b) = a \odot b$ which combines its input with $a$ (on the left). This function is injective, that is, one-to-one: if $f_a(b) = f_a(c)$, then by definition $ab = ac$, and combining both sides with $a^{-1}$ on the left, $a^{-1}ab = a^{-1}ac$, hence $b = c$. (In a group we can always cancel things from both sides of an equation—though only from the end! For example, from $abc = xby$ it does not follow that $ac = xy$.) Conversely, this means that if $b \neq c$, then $ab \neq ac$.

When we form the left coset $aH$, we are applying the function $f_a$ to every element of $H$. The fact that this function is injective means it can’t “collapse” multiple elements of $H$ into the same element in the result. This shows that the coset $aH$ has to have the same size as $H$: there is exactly one element in $aH$ for each element of $H$, and they all have to be different. 