Unable to Understand the meaning of a term in question in book Richard Brualdi Introductory Combinatorics in Chapter Polya Counting

I have used Brualdi's book in the past, but do not currently have access to a copy. Google Books shows me three relevant excerpts from the 2nd edition. The first is on page 554:

As a result a set of permutations which results by considering all the symmetries of a figure is automatically a permutation group. Thus we have a corner-symmetry group, an edge-symmetry group, a face-symmetry group and so on.

Servaes has now posted a very useful answer that includes a longer excerpt containing the same text.

The second is a pair of exercises on page 590:

5. Determine the symmetry group and corner-symmetry group of a triangle which is neither equilateral nor isosceles.

6. Determine the symmetry group of a regular tetrahedron. (Hint: There are 12 symmetries.)

The third, on page 609, is an apparent answer to an exercise:

3. The symmetry group contains only the identity motion. The corner symmetry group contains only the identity permutation of the three corners.

This looks like the answer to Exercise 5 mentioned above, although the number of the problem does not correspond.

The quotation in Servaes's answer confirms that the symmetry group of a figure is the set of geometric motions that map the figure to itself. You will have to find the relevant passages in the book that define whether these motions include rotations and reflections or rotations only. (I would start by looking for the text quoted above. The surrounding text may contain useful information.) My guess, based on the hint to Exercise 6, is that it is rotations only.

I will discuss the tetrahedron because it provides a good example of why it is necessary to distinguish the symmetry group, corner-symmetry group, edge-symmetry group, and face-symmetry group, even though all four groups are isomorphic in this case. Assuming we are concerned with rotations only, the symmetry group of the regular tetrahedron consists of the identity motion, eight rotations of either $120^\circ$ or $240^\circ$ about the four axes joining a corner to the center of the opposite face, and three $180^\circ$ rotations about the axes joining the centers of opposite edges. To fully describe the group, the rules for composing these motions need to be stated. Note that a set of motions forms a group in the algebraic sense when it is closed under composition of motions and contains inverses of all motions.

One typically gives names to each motion. Label the tetrahedron vertices with the numbers $1$ through $4$, with $2$ $3$, and $4$ in counterclockwise order as viewed from vertex $1$. You might use $r_1$ to mean a $120^\circ$ clockwise rotation about the axis through vertex $1$ (as viewed from above). Then the $240^\circ$ rotation about this axis would be $r_1^2$. Similarly, you might use $f_1$ for the $180^\circ$ rotation about the axis through the center of edge $1$-$4$ and the center of edge $2$-$3$. The relations $r_1^3=e$ and $f_1^2=e$ hold, where $e$ is the identity motion (no motion at all). Also $r_1f_1$ (the operation of first doing $r_1$, then doing $f_1$) equals $r_2$ (the $120^\circ$ clockwise rotation about the axis through vertex $2$, viewed from a vantage point above vertex $2$). Two other $120^\circ$ rotations, $r_3$ and $r_4$ are defined similarly, as are two other $180^\circ$ rotations, $f_2$, about the axis joining the centers of edges $2$-$4$ and $1$-$3$, and $f_3$, about the axis going the centers of edges $3$-$4$ and $1$-$2$. Hence the symmetry group is the set of motions $$ \{e,r_1,r_1^2,r_2,r_2^2,r_3,r_3^2,r_4,r_4^2,f_1,f_2,f_3\} $$ with composition of motions the group operation. By a detailed analysis, one can work out the "multiplication table" for the group, which shows how all $12$ motions compose with each other.

The corner-symmetry group would be obtained by writing down the permutations of the vertices that correspond to each of the motions listed above. The permutations corresponding to $r_1$, $f_1$, and $r_2$ are $(23)(24)$, $(14)(23)$, and $(14)(13)$. The symmetry group and the corner symmetry group are clearly isomorphic. The corner-symmetry group acts on a four-element set, the set $\{1,2,3,4\}$ of vertex labels. The symmetry group, in contrast, acts on the geometrical object itself.

One can define the edge symmetry group in a similar way by labelling edges and determining how each of the motions permutes the edges. Let edges $1$-$4$, $2$-$4$, $3$-$4$, $2$-$3$, $1$-$3$, and $1$-$2$ be labeled $a$, $b$, $c$, $x$, $y$, and $z$. Then $r_1$ corresponds to the permutation $(az)(ay)(bx)(bc)$, $f_1$ to the permutation $(by)(cz)$, and $r_2$ to the permutation $(ac)(ay)(bx)(bz)$. The edge-symmetry group acts on a six-element set, the set $\{a,b,c,x,y,z\}$ of edge labels.

Likewise for the face-symmetry group, which acts on the four-element set of face labels. If each face is labeled with the same label as the vertex opposite, the face-symmetry group is identical to the corner-symmetry group and both are identical to the alternating group $A_4$ (consisting of all even permutations) on the set $\{1,2,3,4\}$.

All four groups are, in fact, isomorphic to each other, meaning, for example, that if each element in the multiplication table of the symmetry group is replaced with the corresponding element of the corner-symmetry group, the result is the correct multiplication table for the corner-symmetry group. This raises the question: will the permutation group associated with any set of geometrical features of the tetrahedron will be isomorphic to the symmetry group? The answer is no, as a trivial example shows: the body-symmetry group consists only of the identity motion since the tetrahedron has only one three-dimensional "facet", which always maps to itself. A more interesting example comes from the set of three $180^\circ$ rotation axes mentioned in connection with the rotations $f_1$, $f_2$, and $f_3$. If these axes are called $I$, $J$, and $K$, we can ask how they permute among themselves under the $12$ geometric moves in the symmetry group. You can check that the three $180^\circ$ rotations, as well as the identity motion, all produce the identity permutation. The rotations $r_1$, $r_2$, $r_3$, and $r_4$ all produce the permutation $(IK)(IJ)$, while their squares all produce the permutation $(IJ)(IK)$. These are the cyclic permutations of $I$, $J$, and $K$ and therefore the "$180^\circ$-axis-symmetry group" is isomorphic to the cyclic group of order $3$. There is a homomorphism of the symmetry group of the tetrahedron to this group, but not an isomorphism.

I'll leave it for you to work out the symmetry group and corresponding corner symmetry group for the equilateral triangle. Assuming only rotations are to be considered, this is a group with three elements. The symmetry group, corner symmetry group, and edge symmetry group are all isomorphic in this case, but the face symmetry group is not, as it's the group containing only the identity permutation.

On page 558 of the fifth edition of the book:

Let $\Omega$ be a geometrical figure. A symmetry of $\Omega$ is a (geometrical) motion or congruence that brings the figure $\Omega$ onto itself.

Hence the symmetry group of an equilateral triangle is the group of symmetries that map the equilateral triangle onto itself. Further in the same paragraph:

Hence we conclude that the symmetries of $\Omega$ act as a permutation group $G_C$ on its corners, a permutation group $G_E$ on its edges, and, in the case where $\Omega$ is three-dimensional, a permutation group $G_F$ on its faces. As a result, a set of permutations that results by considering all the symmetries of a figure is automatically a permutation group. Thus, we have a corner-symmetry group, an edge-symmetry group, a face-symmetry group, and so on.

Hence the corner symmetry group is the group of symmetries of the corners obtained by the action of the symmetry group of $\Omega$ on the corners.