Understanding this action from what you said requires a basic understanding of the objects of music theory as members of a set with some structure. I’ll try and give the briefest of explanations possible. The set of *pitch classes* in (Western chromatic) harmony is a twelve-element set, which we will identify with the set of integers modulo 12. The natural cyclic ordering on the set of pitch classes agrees with the cyclic ordering on $C_{12} = \mathbb{Z}/12\mathbb{Z}$, so we will use it. An *interval* is a pair of pitch classes $\{a,b\}$, and its *quality* is the quantity $q(a,b) = |a - b|$ mod 12. A *minor third* is an interval of quality 3, and a *major third* is an interval of quality 4. A *triad* is a triple of pitch classes $(a,b,c)$, where we always assume $a < b < c$ in the cyclic ordering. We are concerned with *major* and *minor* triads. A triad is *major* if $q(a,b) = 4$ and $q(b,c) = 3$, and a triad is *minor* if $q(a,b) = 3$ and $q(b,c) = 4$. (In other words, we always have $q(a,c) = 7$, and we call the triad major or minor depending on the quality of the interval $\{a,b\}$.) In music theory there are other triads, but for this answer I will use *triad* as a shorthand for *major or minor triad*. The set of triads is a 24-element set, for a triad $(a,b,c)$ is completely characterized by its *root note* $a$ and its quality, major or minor. **The action of transposition and inversion** is maybe easier to describe. Consider the group of bijection of $C_{12}$ generated by $$\begin{cases}\iota\colon n \mapsto -n & \mod 12 \\ \tau\colon n \mapsto n+ 1 & \mod 12. \end{cases}$$ Note that $q(\iota(a),\iota(b) = q(a,b)$ and $q(\tau(a),\tau(b)) = q(a,b)$, and that this action is a faithful action of the dihedral group of order 24, which I am fond of calling $D_{12}$ for its natural action on a 12-element set, forgive me. Since this action preserves the quality of intervals, there is a diagonal action of $D_{12}$ on the set of triads. In fact, because we know $\tau$ preserves the cyclic ordering and $\iota$ reverses it, we can describe the action completely as $$\begin{cases} \iota\colon (a,b,c) \mapsto (\iota(c),\iota(b),\iota(a)) \\ \tau\colon (a,b,c) \mapsto (\tau(a),\tau(b),\tau(c)).\end{cases}$$ **The “P/L/R” action** is maybe slightly more annoying for someone with zero familiarity with Western harmony—indeed, I already see some confusion about it in the comments from people with plenty of familiarity! To properly explain where this comes from takes us a little further afield, so let me just give definitions. As we saw above, any triad is completely determined by its root note and its quality. Thus we can parametrize a triad as an element of $C_{12}\times C_2$. For some reason, I would like to think of $C_2$ multiplicatively as the group of units of the ring $\mathbb{Z}$, forgive me. Thus the triad $(a,b,c)$ is sent to $(a,+1)$ if $q(a,b) = 4$ and to $(a,-1)$ if $q(a,b)=3$. The *parallel major/minor* of a triad $(a,t) \in C_{12}\times C_2$ is the triad $P(a,t) = (a,-t)$. Thus $P^2 = 1$, and $P(0,3,7) = (0,4,7)$. The *mediant* of a triad is perhaps best described as follows. If $(a,b,c)$ is a triad, then its *mediant* $L(a,b,c)$ is the triad $(b,c,b+7)$. Thus if $(a,b,c)$ is a major triad, i.e. $(a,b,c) \leadsto (a,+1)$, then the mediant is $(b,-1)$. Conversely if $(a,b,c) \leadsto (a,-1)$, then the mediant is $(b,+1)$. From here you can easily reduce this to a map from $C_{12}\times C_2$ to itself, it just requires cases to state. Note that $L$ does *not* have order two. The *relative major/minor* of a triad is again more easy to describe with music theory than directly as an action on a set, so I will content myself with describing it as $R = L^{-1}$. ____ Unfortunately there is a mistake. You can verify for yourself that the element $L$ has order $24$, and thus the group $\langle P,L,R\rangle$ cannot be the dihedral group of order $24$. Let me just show that $L$ does *not* have order $12$. Recall that $L$ inverts the quality of the triad and translates by $4$ or by $3$ according to the quality of the triad. We see that $L^2$ preserves the quality and has translated both by $4$ and by $3$ and thus $L^2$ satisfies $L^2 = \tau^7$. Thus $L^{12} = \tau^{6\cdot 7} = \tau^{42} = \tau^6$, since $\tau$ has order $12$.