Normal Subgroups

Consider a group $G=(g_1,g_2,\ldots)$ and let $H=(h_1,h_2,\ldots)$ be a sub-group of $G$. As discussed in class, the left coset $G/H$ is not necessarily a group in the sense that the multiplication law $g_1\cdot g_2 = g_3$ in the group $G$ doesn't induce the following multiplication law in the coset:

\begin{align} [g_1] \cdot [g_2] = [g_3] \quad . \end{align}

Recall that $[g]$ refers to an element of the coset that is given by the set of elements that are in the equivalence class of $g\in G$. When does such an operation make sense? When it does, we say that the subgroup $H$ is a normal subgroup of $G$. This leads to the following definition of a normal subgroup:

Definition 1: A subgroup $H$ of a group $G$ is normal if

\begin{align} g\cdot h \cdot g^{-1} \in H\ , \quad \forall\ h \in H\ \textrm{and} \ \forall\ g \in G\ . \end{align}

Definition 2: A subgroup $H$ of a group $G$ is normal if the left coset $G/H$ forms a group.

Exercise: Show that the two definitions are equivalent i.e., definition 1 implies definition 2 and vice versa.


  1. In definition 2, we can replace left coset by a right coset. In other words, if the left coset is a group, then so is the right coset.
  2. Every group has two trivial normal subgroups — the identity subgroup and the full group itself. A group which has no other normal subgroups is called a simple group. Some examples of simple groups are:
      • The cyclic group, ${\mathbb Z}_p$, with $p$ a prime number is a simple group. Check to see what goes wrong when $p$ is not prime.
      • The alternating group, $A_n$ ($n>4$), that is the subgroup of even permutations in $S_n$, is also a simple group. The smallest non-abelian simple group is $A_5$ which has order 60!
      • There exists a classification of finite simple groups. The classification consists of the two examples that we just considered and a few other (infinite) classes and interestingly another 26 groups that don't fit anywhere. These are called the sporadic groups. The largest of them all, is called the Monster_group (20 of the 26 sporadic groups arise as subgroups of the Monster!), and has order $\sim 8 \times 10^{53}$. To those who are keen to know the exact number, here it is:
\begin{eqnarray} 2^{46} \cdot 3^{20} \cdot 5^9 \cdot 7^6 \cdot 11^2 \cdot 13^3 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 41 \cdot 47 \cdot 59 \cdot 71\qquad\\ = 808017424794512875886459904961710757005754368000000000 \end{eqnarray}

Here is a fun talk by Michael Giudici (U. of Western Australia) titled Moonshine and the monster which you can download and read. It is a big pdf file though — 9+ MB.

Exercise: The centre of a group $G$, $Z(G)$, is defined to the set of elements that commute with all elements of $G$. Show that $Z(G)$ is a normal subgroup of $G$.

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