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It's a bit late answer, but there is a nice proof :).
Denote $t = x^3 = y^5$, so $t$ is in center of $G$. Add new symbol $u$ and state that it commutes with other symbols and $u^{-30}=t$, so we obtain new group $E$ isomorphic to $G \times C_{30}$. Now denote $a = u^{10}x, b=u^{6}y, c=u^{15}yx$. It is easy to check that $a^3=b^5=c^2=1$ and $bac=u$ in $E$. Denote $Q_1 = u^{-1}ba, Q_2 =u^{-1}ab$. You can check that $Q_1^2 = Q_2^2 = 1$. The next holds:
$$Q_1Q_2 = u^{-2}b(a^2)b = u^{-2}b(u^{-2}bab)b = u^{-4}b^2ab^2 = u^{-4}b^2a(b^{-1})b^3 = $$ $$= u^{-4}b^2a(u^{-2}aba)b^3 = u^{-6}b^2a^2bab^3 = u^{-6}(b^2a^2)b(b^2a^2)^{-1}$$ Hence $(Q_1Q_2)^5=u^{-30}=t$. But $(Q_2Q_1)^5 = Q_1(Q_1Q_2)^5Q_1 = t$ so $t^2=(Q_1Q_2)^5(Q_2Q_1)^5=1$ in $E$. Clearly, $t^2=1$ should holds in $G$ too.

The idea of this proof is from this article: "Scalar operators equal to the product of unitary roots of the identity operator", Yu. S. Samoilenko, D. Yu. Yakymenko, Ukrainian Mathematical Journal, November 2012, Volume 64, Issue 6, pp 938-947.

It's a bit late answer, but there is a nice proof :).
Denote $t = x^3 = y^5$, so $t$ is in center of $G$. Add new symbol $u$ and state that it commutes with other symbols and $u^{-30}=t$, so we obtain new group $E$ isomorphic to $Ext(C_{30},G)$. Now denote $a = u^{10}x, b=u^{6}y, c=u^{15}yx$. It is easy to check that $a^3=b^5=c^2=1$ and $bac=u$ in $E$. Denote $Q_1 = u^{-1}ba, Q_2 =u^{-1}ab$. You can check that $Q_1^2 = Q_2^2 = 1$. The next holds:
$$Q_1Q_2 = u^{-2}b(a^2)b = u^{-2}b(u^{-2}bab)b = u^{-4}b^2ab^2 = u^{-4}b^2a(b^{-1})b^3 = $$ $$= u^{-4}b^2a(u^{-2}aba)b^3 = u^{-6}b^2a^2bab^3 = u^{-6}(b^2a^2)b(b^2a^2)^{-1}$$ Hence $(Q_1Q_2)^5=u^{-30}=t$. But $(Q_2Q_1)^5 = Q_1(Q_1Q_2)^5Q_1 = t$ so $t^2=(Q_1Q_2)^5(Q_2Q_1)^5=1$ in $E$. Clearly, $t^2=1$ should holds in $G$ too.

The idea of this proof is from this article: "Scalar operators equal to the product of unitary roots of the identity operator", Yu. S. Samoilenko, D. Yu. Yakymenko, Ukrainian Mathematical Journal, November 2012, Volume 64, Issue 6, pp 938-947.

It's a bit late answer, but there is a nice proof :).
Denote $t = x^3 = y^5$, so $t$ is in center of $G$. Add new symbol $u$ and state that it commutes with other symbols and $u^{-30}=t$, so we obtain new group $E$ isomorphic to $G \times C_{30}$. Now denote $a = u^{10}x, b=u^{6}y, c=u^{15}yx$. It is easy to check that $a^3=b^5=c^2=1$ and $bac=u$ in $E$. Denote $Q_1 = u^{-1}ba, Q_2 =u^{-1}ab$. You can check that $Q_1^2 = Q_2^2 = 1$. The next holds:
$$Q_1Q_2 = u^{-2}b(a^2)b = u^{-2}b(u^{-2}bab)b = u^{-4}b^2ab^2 = u^{-4}b^2a(b^{-1})b^3 = $$ $$= u^{-4}b^2a(u^{-2}aba)b^3 = u^{-6}b^2a^2bab^3 = q^{-6}(b^2a^2)b(b^2a^2)^{-1}$$ Hence $(Q_1Q_2)^5=u^{-30}=t$. But $(Q_2Q_1)^5 = Q_1(Q_1Q_2)^5Q_1 = t$ so $t^2=(Q_1Q_2)^5(Q_2Q_1)^5=1$ in $E$. Clearly, $t^2=1$ should holds in $G$ too.

The idea of this proof is from this article: "Scalar operators equal to the product of unitary roots of the identity operator", Yu. S. Samoilenko, D. Yu. Yakymenko, Ukrainian Mathematical Journal, November 2012, Volume 64, Issue 6, pp 938-947.

It's a bit late answer, but there is a nice proof :).
Denote $t = x^3 = y^5$, so $t$ is in center of $G$. Add new symbol $u$ and state that it commutes with other symbols and $u^{-30}=t$, so we obtain new group $E$ isomorphic to $G \times C_{30}$. Now denote $a = u^{10}x, b=u^{6}y, c=u^{15}yx$. It is easy to check that $a^3=b^5=c^2=1$ and $bac=u$ in $E$. Denote $Q_1 = u^{-1}ba, Q_2 =u^{-1}ab$. You can check that $Q_1^2 = Q_2^2 = 1$. The next holds:
$$Q_1Q_2 = u^{-2}b(a^2)b = u^{-2}b(u^{-2}bab)b = u^{-4}b^2ab^2 = u^{-4}b^2a(b^{-1})b^3 = $$ $$= u^{-4}b^2a(u^{-2}aba)b^3 = u^{-6}b^2a^2bab^3 = u^{-6}(b^2a^2)b(b^2a^2)^{-1}$$ Hence $(Q_1Q_2)^5=u^{-30}=t$. But $(Q_2Q_1)^5 = Q_1(Q_1Q_2)^5Q_1 = t$ so $t^2=(Q_1Q_2)^5(Q_2Q_1)^5=1$ in $E$. Clearly, $t^2=1$ should holds in $G$ too.

The idea of this proof is from this article: "Scalar operators equal to the product of unitary roots of the identity operator", Yu. S. Samoilenko, D. Yu. Yakymenko, Ukrainian Mathematical Journal, November 2012, Volume 64, Issue 6, pp 938-947.

It's a bit late answer, but there is a nice prove :).
Denote $t = x^3 = y^5$, so $t$ is in center of $G$. Add new symbol $u$ and state that it commutes with other symbols and $u^{-30}=t$, so we obtain new group $E$ isomorphic to $G \times C_{30}$. Now denote $a = u^{10}x, b=u^{6}y, c=u^{15}yx$. It is easy to check that $a^3=b^5=c^2=1$ and $bac=u$ in $E$. Denote $Q_1 = u^{-1}ba, Q_2 =u^{-1}ab$. You can check that $Q_1^2 = Q_2^2 = 1$. The next holds:
$$Q_1Q_2 = u^{-2}b(a^2)b = u^{-2}b(u^{-2}bab)b = u^{-4}b^2ab^2 = u^{-4}b^2a(b^{-1})b^3 = $$ $$= u^{-4}b^2a(u^{-2}aba)b^3 = u^{-6}b^2a^2bab^3 = q^{-6}(b^2a^2)b(b^2a^2)^{-1}$$ Hence $(Q_1Q_2)^5=u^{-30}=t$. But $(Q_2Q_1)^5 = Q_1(Q_1Q_2)^5Q_1 = t$ so $t^2=(Q_1Q_2)^5(Q_2Q_1)^5=1$ in $E$. Clearly, $t^2=1$ should holds in $G$ too.

The idea of this proof is from this article: "Scalar operators equal to the product of unitary roots of the identity operator", Yu. S. Samoilenko, D. Yu. Yakymenko, Ukrainian Mathematical Journal, November 2012, Volume 64, Issue 6, pp 938-947.

It's a bit late answer, but there is a nice proof :).
Denote $t = x^3 = y^5$, so $t$ is in center of $G$. Add new symbol $u$ and state that it commutes with other symbols and $u^{-30}=t$, so we obtain new group $E$ isomorphic to $G \times C_{30}$. Now denote $a = u^{10}x, b=u^{6}y, c=u^{15}yx$. It is easy to check that $a^3=b^5=c^2=1$ and $bac=u$ in $E$. Denote $Q_1 = u^{-1}ba, Q_2 =u^{-1}ab$. You can check that $Q_1^2 = Q_2^2 = 1$. The next holds:
$$Q_1Q_2 = u^{-2}b(a^2)b = u^{-2}b(u^{-2}bab)b = u^{-4}b^2ab^2 = u^{-4}b^2a(b^{-1})b^3 = $$ $$= u^{-4}b^2a(u^{-2}aba)b^3 = u^{-6}b^2a^2bab^3 = q^{-6}(b^2a^2)b(b^2a^2)^{-1}$$ Hence $(Q_1Q_2)^5=u^{-30}=t$. But $(Q_2Q_1)^5 = Q_1(Q_1Q_2)^5Q_1 = t$ so $t^2=(Q_1Q_2)^5(Q_2Q_1)^5=1$ in $E$. Clearly, $t^2=1$ should holds in $G$ too.

The idea of this proof is from this article: "Scalar operators equal to the product of unitary roots of the identity operator", Yu. S. Samoilenko, D. Yu. Yakymenko, Ukrainian Mathematical Journal, November 2012, Volume 64, Issue 6, pp 938-947.