What does "contain full conjugacy classes" mean (in simple English please)? I'm trying to understand the solution to question part $(vi.)$ below:


A subgroup is invariant if $gHg^{−1} = H$ for any $g \in G$. This is equivalent to saying that
a subgroup $H$ is invariant if it has the same right and left cosets, or $\color{red}{\text{the same as saying that it should contain full conjugacy classes}}$. This is true for $\{E\}$, $\{E, D, F \}$ and $G$ which contain full classes, but not for $\{E, A\}$, $\{E, B\}$ and $\{E, C\}$. Also since $\{E, D, F\}$ has index two $\big(\mid G\mid/\mid H\mid = 2 \big)$, from question 2 and my previous question we know that this subgroup is invariant.
As a check, let’s see if $\{E, A\}$ has the same left and right cosets:
Left cosets of $\{E, A\} \,\text{are}\, A\{E, A\} = \{A, E\},\quad B\{E, A\} = \{B, F \},\quad C\{E, A\} = \{C, D\}$
Right cosets of $\{E, A\}\,\text{are}\, \{E, A\}A = \{A, E\},\quad \{E, A\}B = \{B, D\},\quad \{E, A\}C = \{C, F\}$
We see that the left cosets are not the same as the right cosets. Hence $\{E, A\}$ is not an invariant subgroup. The same thing holds for $\{E, B\}$ and $\{E, C\}$.
Therefore only $\{E\}, \color{blue}{\{E, D, F \}}$ and $G$ are invariant subgroups.

I marked the part in red for which I don't understand. In part $iii.$ it was found that the conjugacy classes are $\{E\}$, $\{A,B,C\}$, $\{D,F\}$. So (from the quote above - the last part of the solution to $vii.$), $\color{blue}{\{E,D,F\}}$ 'contains full conjugacy classes', but what does this mean in simple English (where possible)?
 A: To see that $\{E,D,F\}$ is a union of whole conjugacy classes, here is the (tedious) work that you must carry out:

*

*The conjugacy class of $E$, namely the set $\{AEA^{-1},BEB^{-1},CEC^{-1},DED^{-1},EEE^{-1},FEF^{-1}\}$, is equal to $\{E\}$ (verify by tedious calculation, or more easily by noticing that $E$ is the identity element of the group).

*The conjugacy class of $D$, namely the set $\{ADA^{-1}, BDB^{-1}, CDC^{-1}, DDD^{-1}, EDE^{-1}, FDF^{-1}\}$, is equal to $\{D,F\}$ (verify by tedious calculation).

*The conjugacy class of $F$ is also equal to $\{D,F\}$ (verify using the theorem that the conjugacy classes of any group are a partition of that group, and the fact that $\{D,F\}$ is a conjugacy class which you have already verified and which contains $F$).

Therefore, the set $\{E,D,F\}$ is a union of two whole conjugacy classes, namely
$$\{E,D,F\} = \{E\} \cup \{D,F\}
$$
Notice, there is another whole conjugacy classes that is not in this union. So, to say that a subset $X$ of the group is "a union of whole conjugacy classes" means, more precisely, "there exists a subset of the set of all conjugacy classes whose union is $X$".
Or, to put it another way, "$X$ is the union of the conjugacy classes of the elements of $X$". Notice: no element of the conjugacy class $\{A,B,C\}$ is contained in the set $X = \{E,D,F\}$. But that does not contradict the meaning of $X$ being a union of whole conjugacy classes: $X$ is not the union of every whole conjugacy class, but it is a union of some of the conjugacy classes.
