Prove that the homophony group is trivial.
{.proof} definition of the homophony group.
<1>2. “knight” and “night” are homophones, so \(k = e\).
<1>3. “write” and “right” are homophones, so \(w = e\).
<1>4. “see” and “sea” are homophones, and “sea” and “c” are homophones, so \(s = e\) and \(c = e\).
<1>5. “you” and “u” are homophones, so \(y \cdot o = e\).
<1>6. “eye” and “i” are homophones, so \(y = i\).
<1>7. “why” and “y” are homophones, so \(h = e\).
<1>8. “are” and “r” are homophones, so \(a = e\).
<1>9. “queue” and “q” are homophones, so \(u = e\).
<1>10. “jay” and “j” are homophones, so \(y = e\).
<1>11. “ell” and “l” are homophones, so \(l = e\); “eff” and “f” are homophones, so \(f = e\).
<1>12. The remaining letters are trivial by their letter-names: “bee” \(= b\), “tea” \(= t\), “gee” \(= g\), “pee” \(= p\), “vee” \(= v\), “ex” \(= x\), “zee” \(= z\), “em” \(= m\), “en” \(= n\), “oh” \(= o\), “double-u” \(= w\), “aitch” \(= h\).
<1>13. Hence every generator of \(H\) is trivial, so \(H\) is the trivial group.
<1>14. Q.E.D.
Solution
Let \(H\) be the quotient of the free group on the \(26\) letters by the relations \(w_1=w_2\) whenever the two English words \(w_1,w_2\) are homophones. We show each letter is the identity. The elimination below is the classical proof of Mestre–Schoof–Washington–Zagier; every step is an ordinary left/right cancellation in the group after previously eliminated letters are set equal to \(1\).
1. First eliminate the vowels and semivowels\[ e,a,i,o,u,w,y. \]
The homophone relation \(\text{bye}=\text{by}\) gives \[ bye=by\quad\Longrightarrow\quad e=1 \] by left cancellation of the common prefix \(by\). Then \[ \begin{array}{rcl} \text{lead}=\text{led}&\Longrightarrow&a=1,\\ \text{maid}=\text{made}&\Longrightarrow&i=1,\\ \text{sow}=\text{sew}&\Longrightarrow&o=1,\\ \text{buy}=\text{by}&\Longrightarrow&u=1,\\ \text{sow}=\text{so}&\Longrightarrow&w=1,\\ \text{lye}=\text{lie}&\Longrightarrow&y=1. \end{array} \] For example, after \(e=1\), the first relation becomes \(lad=ld\), so cancellation of \(l\) on the left and \(d\) on the right gives \(a=1\). The others are identical cancellation arguments using letters already eliminated earlier in the displayed order.
2. Next eliminate\[ h,k,n,p,b. \]
Using the relations \[ \text{hour}=\text{our},\qquad \text{knight}=\text{night},\qquad \text{damn}=\text{dam},\qquad \text{psalter}=\text{salter},\qquad \text{plumb}=\text{plum}, \] we obtain respectively \(h=k=n=p=b=1\). Each equality has the form \(u\ell v=uv\) after the letters from 1 and the previously eliminated letters are removed, so cancellation gives \(\ell=1\).
3. Next eliminate\[ s,t,l,r,m. \]
The homophone relations \[ \text{bass}=\text{base},\quad \text{butt}=\text{but},\quad \text{tolled}=\text{told},\quad \text{barred}=\text{bard},\quad \text{dammed}=\text{damned} \] reduce, using the letters already known to be trivial, to \[ s^2=s,\qquad t^2=t,\qquad l^2=l,\qquad r^2=r,\qquad m^2=m. \] In a group an idempotent element is the identity: \(x^2=x\) implies \(x=1\) after multiplying by \(x^{-1}\). Hence \(s=t=l=r=m=1\).
4. Eliminate \(d\) and \(g\).
The relations \[ \text{chased}=\text{chaste},\qquad \text{sign}=\text{sine} \] reduce, after the previous eliminations, to \(d=1\) and \(g=1\) respectively.
5. Eliminate\[ z,c,j,q,x. \]
The homophone relations \[ \text{daze}=\text{days},\qquad \text{cite}=\text{sight},\qquad \text{jeans}=\text{genes},\qquad \text{queue}=\text{cue},\qquad \text{tax}=\text{tacks} \] reduce successively to \[ z=1,\qquad c=1,\qquad j=1,\qquad q=1,\qquad x=1, \] using only letters already eliminated before the relevant step.
6. Finally eliminate \(f\) and \(v\).
The relation \(\text{phase}=\text{faze}\) reduces to \(f=1\) because \(p,h,a,s,e,z\) are already trivial. Then \[ \text{chivvy}=\text{chivy} \] reduces to \(v^2=v\), hence \(v=1\).
7. Therefore every one of the \(26\) generators is trivial, so\[ \boxed{H=1}. \]
The letters eliminated in <1>1–<1>6 are \[ a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z, \] which is the complete generating set.
the relation \(\text{knight} = \text{night}\) cancels the common suffix, leaving \(k = e\).
cancelling the common “rite” leaves \(w = e\).
\(\text{see} = \text{sea}\) gives \(e = a\) (after cancelling \(s\) and \(e\)), and \(\text{sea} = c\) gives \(s \cdot e \cdot a = c\), so \(s = e\) and \(c = e\).
\(\text{you} = u\) cancels the trailing \(u\), leaving \(y \cdot o = e\).
\(\text{eye} = i\) gives \(e \cdot y \cdot e = i\), i.e. \(y = i\).
\(\text{why} = y\) gives \(w \cdot h \cdot y = y\), so \(w \cdot h = e\), and \(w = e\) by 3, hence \(h = e\).
\(\text{are} = r\) gives \(a \cdot r \cdot e = r\), so \(a = e\).
\(\text{queue} = q\) gives \(q \cdot u \cdot e \cdot u \cdot e = q\), so \(u = e\).
\(\text{jay} = j\) gives \(j \cdot a \cdot y = j\), so \(a \cdot y = e\), and \(a = e\) by 8, hence \(y = e\).
\(\text{ell} = l\) gives \(e \cdot l \cdot l = l\), so \(l = e\); \(\text{eff} = f\) gives \(e \cdot f \cdot f = f\), so \(f = e\).
each letter-name is a homophone of the letter itself, and the extra letters in each name are already trivial by <1>2–<1>11.
<1>2–<1>12 cover all \(26\) letters.
13.