In group theory, in order to assign a point group, one must first identify the symmetry elements present in the molecule.

To give an example, water has:

  • a rotational axis, labelled C2 since it is a two fold axis, with a π/2 rotation giving back (what appears to be)the same thing
  • two mirror planes, labelled σv and σv' (with additional labels xy/xz to define what plane the mirror plane lies in)

Symmetry elements present in water

Symmetry elements present in water

(Note that for the C2v point group, the axis are defined 'arbitrarily', different tables/books might swap x and y)

From this, it would look like water has three unique symmetry elements, however this is not the case, in fact, water has four due to the presence of an additional symmetry element, E, known as the identity operator (sometimes seen as I in certain textbooks).

This E operator corresponds to 'do nothing' (i.e. leave atoms where they are), which is easily seen in the character table for the C2v point group (to which water belongs) in which the characters under E are all '1':

$$\begin{array}{c|cccc|cc} \hline C_\mathrm{2v} & \color{red}{E} & C_2 & \sigma_\mathrm{v}(xz) & \sigma_\mathrm{v}'(yz) & & \\ \hline \mathrm{A_1} & \color{red}{1} & 1 & 1 & 1 & z & x^2, y^2, z^2 \\ \mathrm{A_2} & \color{red}{1} & 1 & -1 & -1 & R_z & xy \\ \mathrm{B_1} & \color{red}{1} & -1 & 1 & -1 & x, R_y & xz \\ \mathrm{B_2} & \color{red}{1} & -1 & -1 & 1 & y, R_x & yz \\ \hline \end{array}$$


C2v character table, from orthocresol's Group Theory Tables

The identity, E, consists of doing nothing; the corresponding symmetry element is the entire object. Because every molecule is indistinguishable from itself if nothing is done to it, every object possesses at least the identity element. (Taken from Atkins' Physical Chemistry).

Since transforming a molecule by the identity operator just gives back itself, the use of 'E' in a qualitative sense appears redundant, however it appears in every character table, and as such much have a purpose.

Why do we need the identity operator, E, and what use does it have in chemical group-theory?

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    $\begingroup$ Not enough to be a full answer, but: The "why do we need" part has been covered thoroughly. As for "what use it has", to state the obvious, the character of an irrep under $E$ is the dimensionality of the irrep. $\endgroup$ Commented Jul 6, 2017 at 2:27
  • $\begingroup$ @orthocresol, good point, aaide for being needed to complete the riteria of s group I'd somewhat forgotten that there was a consequence when actually using group theory. $\endgroup$
    – NotEvans.
    Commented Jul 6, 2017 at 6:11
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    $\begingroup$ What element of your group is $C_2^4$? Groups are closed under multiplication, so any product of elements of the group is some element of the group... $\endgroup$ Commented Jul 6, 2017 at 11:02
  • $\begingroup$ @EricTowers, could you expand, not entirely sure what you're saying/what point you're making $\endgroup$
    – NotEvans.
    Commented Jul 6, 2017 at 12:26
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    $\begingroup$ You say $C_2$ is an element of your group of symmetries. This forces $C_2 \cdot C_2 \cdot C_2 \cdot C_2$ to also be a symmetry. Which element is that product? (Using the fourth power is a consequence of an error in your description of $C_2$. Actually, $C_2$ is rotation by $\pi$ radians. With this correction, we need only ask "which symmetry is given by 'first do $C_2$, then do $C_2$ again'?") $\endgroup$ Commented Jul 6, 2017 at 16:39

6 Answers 6


It is not that we need the identity operator. It is just that things (character tables, irreducible representations, etc.) work the way they do. As to why they do so, I refer you to any decent textbook on group theory.

I guess one could rebuild the entire theory from scratch (that is, reformulate all definitions, starting from the very definition of a group, reprove all theorems, etc.) without explicitly using $\rm E$. That would be a titanic amount of work, though, and quite pointless at that. Also, many things like "order of a group is divisible by..." would start to sound less natural.

Come to think of it, why do we need the number $0$ in math? It just sits there and does nothing, after all. Well, doing nothing may be quite important: it makes a lot of other things simpler.

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    $\begingroup$ ♫ Zero is the loneliest number♫ $\endgroup$
    – Cort Ammon
    Commented Jul 5, 2017 at 23:06
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    $\begingroup$ Of course, zero is the identity element in the group consisting of all integers under the operation of addition. $\endgroup$ Commented Jul 6, 2017 at 2:14
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    $\begingroup$ For clarity on why the identity element is so important to group theory in general, it might be worth giving an example of a group and its identity where the identity element isn't so boring and simple as '0' or 'do nothing' such as those in matrix operations and the very complex identity elements in sand pile groups plus.google.com/101584889282878921052/posts/QezmLcTCTMJ $\endgroup$ Commented Jul 6, 2017 at 18:28
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    $\begingroup$ For a set $G$ to be a group, every $a \in G$ must have an inverse $a^{-1}$ which is defined by satisfying $a \times a^{-1} = a^{-1} \times a = e$, where $e$ is the identity of the group. It is thus impossible to abolish $e$. Same thing happens for the integers and its zero. $\endgroup$ Commented Jul 6, 2017 at 22:35
  • $\begingroup$ Put another way, if there weren't an identity, we wouldn't be talking about a rotation group; we'd just be talking about a rotation semigroup :) $\endgroup$
    – chepner
    Commented Jul 7, 2017 at 3:22

What you are asking arises as a fundamental consequence of the definition of vector spaces and the operations defined for them and, specifically, symmetry groups. The identity operator does not do "nothing" (see below) and it does belong in the formal definition of a group, and it is included to satisfy said definition.

From Byron and Fuller's Mathematics of Classical and Quantum Physics:

...a group is a system consisting of a set of elements and rules for combining the members of this set.

This sets the stage for the definition:

A group is a system $\{G, \cdot\}$, consisting of the set $G$ and a single closed operation $\cdot$, which satisfies the following three axioms:

  1. If $a, b, c \in G$, then $a\cdot(b\cdot c) = (a\cdot b) \cdot c$ (associativity).

  2. There exists an identity element $e\in G$ such that for all $a\in G$, $a\cdot e = e\cdot a = a$.

  3. For every $a\in G$, there exists an inverse element in $G$, denoted $a^{-1}$, such that $a\cdot a^{-1} = a^{-1}\cdot a = e$.

Given this, we can go to more plain English as Wilson, Decius, and Cross do in Molecular Vibrations, where they note that

It is evident that for every symmetry operation possessed by a molecule there is some operation, either the same one or a different one, which undoes the work of the first.

This is an important point, and operationally illustrates what the identity operator does and why it is included in every group:

...if an operation $R$ changes a molecule from the position I to the position II, then there is an operation which shifts the molecule from II to I. Such an operation is called the inverse of the original operation and is written $R^{-1}$. Symbolically, $R^{-1}R = E$, since the symbol $E$ represents the identity operation, which leaves the molecule unaltered.

(For the record, WDC also write that "...$E$ is used for the trivial operation of leaving a molecule in its original position.")

  • $\begingroup$ I learnt that there were four axioms (the fourth being that for all $a, b \in G$, $a \cdot b$ must also be in $G$), but it's been a while. Maybe it's already implied by the other three axioms, I'm not sure. $\endgroup$ Commented Jul 6, 2017 at 6:22
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    $\begingroup$ I am sorry, I must have read this in a hurry; as my "fourth axiom" is precisely equivalent to the phrase "closed operation" in Todd's quote. // I assume the ultimate authority on the matter to be a maths textbook. Looking through my notes from Rotman's Abstract Algebra it does list only three axioms. However, like Todd's source, Rotman also separately defines the binary operation, $*$, to obey $G * G \to G$. Essentially, that means $*$ takes two arguments, both in the set $G$, and outputs one value which is also in the set $G$. [cc @NotEvans.] $\endgroup$ Commented Jul 6, 2017 at 9:02
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    $\begingroup$ @orthocresol I think it ultimately comes down to whether you think of the "closedness" is a property of the operation or a property of the set. There was an interesting discussion about this on the talk page for the Wikipedia article on (algebraic) magmas. The proponents of the view that it's a property of the operation, pointed out it follows naturally from the definition of a binary operation as being from $G \times G \to G$ (as in Rotman). But I've certainly seen both. $\endgroup$
    – Silverfish
    Commented Jul 6, 2017 at 11:00
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    $\begingroup$ For clarity on why the identity element is so important to group theory in general, it might be worth giving an example of a group and its identity where the identity element isn't so boring and simple as '0' or 'do nothing' such as those in matrix operations and the very complex identity elements in sand pile groups plus.google.com/101584889282878921052/posts/QezmLcTCTMJ $\endgroup$ Commented Jul 6, 2017 at 18:27

Well, according to this Math StackExchange post, you need the identity element because it is impossible to define an inverse element a if you don't know about the identity element e. The definition of a group includes the identity element (below).

There exists an element e in [Group] G such that, for every element a in G, the equation e • a = a • e = a holds. Such an element is unique (see below), and thus one speaks of the identity element. (Wikipedia)


Symmetry operator $E$ is the lowest symmetry a molecule may possess; and indeed all objects possess this operator. Yet, among molecules revealing $E$ as the only the operation are the ones exhibiting stereogenic centred chirality, like $\ce{CFClBrI}$. Hence $E$ is of value, not redundant, and needed.

  • $\begingroup$ Actually, as far as I know, electrons do not possess it. Spinor is identical after 720 deg. rotation, not 360. $\endgroup$
    – Mithoron
    Commented May 29, 2021 at 13:05

This is a special case of a phenomenon that occurs throughout mathematics. For example, you could ask why we have zero for similar reasons. The answer is frequently: it's often easier to reason about a collection of things if the thing that represents 'nothing' is considered part of the collection.

For example, if we didn't have zero there'd be two kind of sentences about how much money we have: "I have \$X" or "I don't have any money". If you allow X to be zero then you only need one kind of sentence "I have \$X".

In the case of groups, many theorems are easier to state if you include the identity as part of your collection. For example "the order of a subgroup divides the order of a group" is much easier than "one plus the order of a subgroup divides one plus the order of a group". And "in a group you can always multiply two elements" becomes "you can multiply elements most of the time but there are certain pairs you can't multiply because when you multiply them they cancel out doing nothing but we don't consider doing nothing to be an element of the group". Closer to your example, the statement of the orthogonality relations for character tables would be more complex.

The fact that so many things simplify when you include the identity is good reason for it being there.


It is true that the identity operation belongs to every group, as the other answers point out. One can still ask, why don't we just omit the first column, by convention, since it is all ones anyway?

The simple answer is that it is not always all-ones. It gives important information, namely the dimension of the space of an irreducible representation. In other words, if the first entry is $d$, the group elements are represented by orthogonal $d\times d$ matrices.

(A less serious reason is that the character table would no longer be a square table if we omit the first column.)


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