Question

First member of a group differs from the rest of the members of the same group. Why?

Answer 1

I interpreted this to be Group Theory... of course, it's up in the air, and could be about the periodic table.

However, elements in a single periodic table group should be quite similar, except for mainly their atomic number, and possibly their phase at `25^@ "C"` and `"1 atm"`. Otherwise, this periodicity of chemical properties is a significant reason for why the periodic table is even arranged in the way it is in the modern day.

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The four postulates of a group are (given a group operation `@`):

1. There exists an identity element `E` in the group, such that for a given other element `A` in the group, `A@E = E@A = A`.
2. There exists an inverse element `B` in the group such that `A@B = B@A = E`.
3. The group operation `@` is associative, i.e. `A@(B@C) = (A@B)@C = A@B@C`.
4. Any two members `A` and `D` in the group via the operation `A@D` generate another element `F` also in the group. We call this the "closure property".

For example, water belongs to the point group `C_(2v)`, and the elements in that group are `E` (identity), `C_2(z)` (two-fold rotation), `sigma_v(xz)` (mirror plane), and `sigma_v(yz)` (mirror plane).

The first element in a group is generally considered the identity element `E`, which differs from the rest because it is its own inverse. An element whose inverse is itself belongs to its own class.

The previous statement can be shown by performing a similarity transform on `E`. This says that:

If `P^(-1)@Q@P = R`, then `R` is conjugate with `Q`, meaning it is in the same class as `Q`. Thus, if `R = Q`, then `Q` is in its own class.

And using the first postulate of a group,

`P^(-1)@(E@P) = P^(-1)@P = E`
`P^(-1)@(E@P) = (P^(-1)@P)@E = E@E = E`

Either way you approach the similarity transformation, `E` is in its own class and thus is nothing like any other element in the group.