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

Sep 16, 2017

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 ${25}^{\circ} \text{C}$ and $\text{1 atm}$. 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.

The four postulates of a group are (given a group operation $\circ$):

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

For example, water belongs to the point group ${C}_{2 v}$, and the elements in that group are $E$ (identity), ${C}_{2} \left(z\right)$ (two-fold rotation), ${\sigma}_{v} \left(x z\right)$ (mirror plane), and ${\sigma}_{v} \left(y z\right)$ (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 ${P}^{- 1} \circ Q \circ P = R$, 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,

${P}^{- 1} \circ \left(E \circ P\right) = {P}^{- 1} \circ P = E$
${P}^{- 1} \circ \left(E \circ P\right) = \left({P}^{- 1} \circ P\right) \circ E = E \circ E = E$

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