# What's the relation between two symmetry groups, if one has all the symmetry of the other and some more?

#### Question:

Consider two molecular symmetry groups, for example $$C_s$$ and $$C_{2v}$$.

• $$C_s$$ has one inversion plane, and two irreductible representations: the symmetric $$A'$$, and the antisymmetric $$A''$$.
• $$C_{2v}$$ has one inversion plane and one 2-fold axis; and four irreductible representations: $$A_1$$, $$B_1$$, $$A_2$$, $$B_2$$.

It is clear that if a molecule has a $$C_{2v}$$ symmetry, it also has a $$C_s$$ symmetry. It also stands (in a selected coordinate system), that an orbital of a molecule with $$C_{2v}$$ symmetry that belong to $$A_1$$ or $$B_1$$ also belong to $$A'$$, and similarly, orbitals belonging to $$A_2$$ and $$B_2$$ also belong to $$A''$$.

1. What mathematical term describes the relationship between $$C_s$$ and $$C_{2v}$$? Is $$C_{2v}$$ a subgroup of $$C_s$$? Is it rather a subset of it? Should we rather say that $$C_{2v}$$ implies $$C_s$$?
2. What's the correct term for the relationship between $$A_1$$/$$B_1$$ and $$A'$$, and between $$A_2$$/$$B_2$$ and $$A''$$? Are they subspecies? Does $$A_1$$ imply $$A'$$?

#### Background:

I'm comparing some orbitals of two similar molecules. One is a root with $$C_{2v}$$ symmetry, and the other is an alkylated version of it with $$C_s$$ symmetry. For certain reasons, symmetry classifications of the molecules play an important part of the writeup. I want to explicitly point out that the symmetry and the irreductable representations of the root correspond to the ones of the alkylated molecule, even though they're labeled differently.

While this sounds like a basic question, I found it surprisingly hard to find an answer. I've been struggling for almost a week at this point. All textbooks and online resources talk about how you can obtain $$C_{2v}$$ from $$C_s$$ with certain symmetry operations, and about what properties they have individually, but never mention the terms I'm looking for.

I don't really have a strong background in group theory, so if I used any other terms incorrectly, I'm glad to be corrected.

• With groups, subgroup is indeed the correct word. Smaller group is a subgroup of the bigger one. Mar 22 at 10:59
• I think you mean "irreducible". I'm not aware of "irreductible" being used, but I could be unaware of its use. "orbitals belonging to A2 and B2 also belong to A′′" My understanding is that elements of A2, B2, and A'' represent symmetries of the molecules. Aren't the symmetries of the molecules different from orbitals? "One is a root with C_{2v} symmetry, and the other is an alkylated version of it with C_s symmetry." Alkylating makes molecules more symmetric? Mar 23 at 0:06
• @Acccumulation 1) $C_{2v}$ is more symmetric than $C_s$ as it also has a 2-fold axis in addition to the inversion plane. 2) Irreps are not the same as symmetries. Vaguely (in this case) they classify whether the orbitals are symmetric or antisymmetric WRT to the specific symmetry operations of the group, but u/orthocresol is much more equipped to comment on this. 3) re spelling, I'll check, it may be a typo that sticks around in my team. Mar 23 at 9:59
• If the user who downvoted the question could gently provide an explanation, I'd be more than glad to attempt to improve it. Right now I'm unsure what's the issue. Mar 23 at 10:00
• @uhoh Probably I used the word wrongly. English is not my native, and until now I believed "disingenious" is pretty much a slightly stronger but still formal and uncharged synonym of "unintuitive". I've seen it a couple times on SE in contexts which led me to this interpretation. Thanks for clearing that up. Mar 27 at 9:23

1. What mathematical term describes the relationship between $$C_s$$ and $$C_{2v}$$? Is $$C_{2v}$$ a subgroup of $$C_s$$? Is it rather a subset of it? Should we rather say that $$C_{2v}$$ implies $$C_s$$?

$$C_\mathrm{s}$$ is a subgroup of $$C_\mathrm{2v}$$, yes.

I'm speaking very informally here, but a group is a set of elements, together with a binary operator $$*$$ that 'combines' the elements to form other elements. In the case of symmetry groups, these elements are symmetry operations. The binary operator is composition, i.e. $$B * A$$ means do symmetry operation $$A$$ first and then $$B$$ (or the other way round, it's just a matter of definition); and it turns out that $$B * A$$ is always also a symmetry element. So, a group has more structure than just a set. A set is just a collection of elements, it doesn't have to have a way to combine them.

It's true that the elements of $$C_\mathrm{s}$$ are a subset of the elements of $$C_\mathrm{2v}$$. However, when referring to the groups themselves, subgroup is the correct term.

I don't think 'implies' is appropriate here, that's typically used in a context like propositional logic.

1. What's the correct term for the relationship between $$A_1$$/$$B_1$$ and $$A'$$, and between $$A_2$$/$$B_2$$ and $$A''$$? Are they subspecies? Does $$A_1$$ imply $$A'$$?

$$A$$, $$B$$, etc. are representations of a group. Defining the word 'representation' is actually quite involved! I've written a bit about it here before, but it's a whirlwind tour, and I don't necessarily think it makes for good reading.

If you're just looking for the correct term, though, then my understanding is that $$A'$$ is the restriction of the representation $$A_1$$ to the subgroup $$C_\mathrm{s}$$: https://en.wikipedia.org/wiki/Restricted_representation

Writing this in a chemistry paper/report will probably throw most readers off. (I reckon most chemists probably don't understand the formal meaning of a subgroup, but would have an intuitive understanding of it. I don't think a restricted representation means anything to the typical chemist, though, except for the very mathematically inclined.) So if I were in your position, I'd probably write that there is a 'correspondence' or a 'correlation' between the two irreps. That's not super-precise language, in that it doesn't describe the nature of the 'correspondence', but it'll get the message across. (Of course, you could define it properly once at the beginning, and then stick to vague language afterwards.)

For a different perspective, these kinds of relationships between representations are tabulated in so-called descent in symmetry tables, which you can find online. I also spent some time typing them up on Chemistry Meta, and in the original source from Atkins et al., similarly vague language is used:

The following tables show the correlation between the irreducible representations of a group and those of some of its subgroups. In a number of cases more than one correlation exists between groups.

• "I also spent some time..." seems like a bit of an understatement :-)
– uhoh
Mar 22 at 18:13

As the discussion of these issues in the comments has gotten rather extensive, I think that I should move it to an answer.

In math, it's common to use the word "symmetry" to refer to an automorphism that leaves an object essentially the same. For example, see this page that says that a cube has 48 symmetries. Thus, what you refer to as a "symmetry operation" would in math often be referred to as simply a "symmetry".

As another example, a benzene molecule, if we ignore the single/double bond issue, has 12 symmetries, six rotations and six reflections. This is known as the dihedral group (note the statement "A regular polygon with n sides has 2n different symmetries", again using the word "symmetry" to refer to an automorphism) $$D_6$$. There are 12 symmetries in $$D_6$$. $$D_6$$ is a group of symmetries.

If we alkylate one of the carbon atoms (don't worry if this doesn't make sense chemically, I'm just dealing with the math), then the molecule has only one non-trivial symmetry (a reflection about the alkylated carbon). This is the binary or $$Z_2$$ group. If we alkylate two carbon atoms at opposite side of the molecule, then there are four symmetries, all of which are members of $$D_6$$. Since all symmetries of alkylated molecules are members of $$D_6$$, it would be correct to say "If the alkylated molecule has a symmetry, then the root has that symmetry".

Note that there is only one way to have a non-alkylated benzene molecule, but many different ways to have an alkylated benzene molecule. So:

• The root has more symmetries than the alkylated molecule.
• For each symmetry, if the alkylated molecule has that symmetry, then the root does as well. For instance, if a molecule has a $$Z_2$$ symmetry, then it has a $$D_6$$ symmetry.
• The root is more symmetric than the alkylated molecule.
• There are more alkylated molecules than root molecules [at least, in terms of possible configurations, not necessarily instances].
• The symmetry group of the alkylated molecule is a subgroup of the root's.

Note that there is an opposite relationship between the first and fourth of these points. If the symmetry group of things of type X is a subgroup of the symmetry group of things of type Y, then the set of things of type Y is a subset of the set of things of type X.

For your question "Is $$C_{2v}$$ a subgroup of $$C_s$$?" the answer is a definite "no"; $$C_s$$ is a subgroup of $$C_{2v}$$. Your statement "It is clear that if a molecule has a $$C_{2v}$$ symmetry, it also has a $$C_s$$ symmetry." is misleading at best. It would be more clear as "It is clear that if a molecule has is invariant under $$C_{2v}$$, then it is invariant under $$C_s$$."