Let $G$ be defined by its group presentation: :$G = \gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$ Let $z$ denote $x^3$. Then: {{begin-eqn}} {{eqn | l = y z y^{-1}       | r = y x^3 y^{-1}       | c =  }} {{eqn | r = \paren {y x y^{-1} }^3       | c = Power of Conjugate equals Conjugate of Power }} {{eqn | r = \paren {x^{11} }^3       | c = from Group Presentation }} {{eqn | r = x^{33}       | c = Powers of Group Elements: Product of Indices }} {{eqn | r = x^{15} x^{15} x^3       | c = Powers of Group Elements: Sum of Indices }} {{eqn | r = e \cdot e \cdot x^3       | c = from Group Presentation }} {{eqn | r = x^3       | c = {{GroupAxiom|2}} }} {{eqn | r = z       | c = Definition of $z$ }} {{eqn | ll= \leadsto       | l = y z       | r = z y       | c = Product of both sides with $y$ }} {{end-eqn}} So $z$ commutes with $y$. As $z$ is a power of $x$, $z$ also commutes with $x$. Hence by definition of center: :$z \in \map Z G$ It follows that $\gen z$ is a normal subgroup of order $5$. Let $K$ be the subgroup of $G$ generated by $x^5$ and $y$. Note that: {{begin-eqn}} {{eqn | l = y x^5 y^{-1}       | r = \paren {y x y^{-1} }^5       | c = Power of Conjugate equals Conjugate of Power }} {{eqn | r = \paren {x^{11} }^5       | c = from Group Presentation }} {{eqn | r = x^{55}       | c = Powers of Group Elements: Product of Indices }} {{eqn | r = x^{15} x^{15} x^{15} x^{10}       | c = Powers of Group Elements: Sum of Indices }} {{eqn | r = e \cdot e \cdot e \cdot x^{10}       | c = from Group Presentation }} {{eqn | r = x^{10}       | c = {{GroupAxiom|2}} }} {{eqn | r = x^{-5}       | c = as $x^{15} = e$ }} {{end-eqn}} Hence the generator of $K$ satisfies: :$\paren{x^5}^3 = e = y^2$ and: :$y x^5 y^{-1} = x^{-5}$ Let $w := x^5$. Then $K$ is generated by $w$ and $y$ where: :$w^3 = 1 = y^2$ and: :$w y = y w^2 = y w^{-1}$ and it is seen that $K$ is isomorphic to the dihedral group $D_3$. It is now to be shown that $G$ is an internal group direct product of $K$ and $\gen z$. We have that $K \cap \gen z = \set e$. From the Internal Direct Product Theorem, we need to prove only that $K$ is a normal subgroup of $G$. We have that: :$\order K = 6$ where $\order K$ denotes the order of $K$. We also have that $K$ is a subgroup of its normalizer $\map {N_G} K$. Hence by Lagrange's Theorem: :$6 \divides \order {\map {N_G} K}$ where $\divides$ denotes divisibility. Again by Lagrange's Theorem: :$\order {\map {N_G} K} \divides 30$ We have: {{begin-eqn}} {{eqn | l = x w x^{-1}       | r = x x^5 x^{-1}       | c = Definition of $w$ }} {{eqn | r = x^5       | c =  }} {{eqn | r = w       | c =  }} {{end-eqn}} demonstrating that $x$ is conjugate to $w$. Then: {{begin-eqn}} {{eqn | l = x y x^{-1}       | r = x y^{-1} x^{-1}       | c = as $y^2 = e$ }} {{eqn | r = y x^{11} x^{-1}       | c = from Group Presentation: $y x y^{-1} = x^{11}$ }} {{eqn | r = y x^{10}       | c =  }} {{eqn | r = y w^2       | c =  }} {{eqn | o = \in       | r = K       | c =  }} {{end-eqn}} demonstrating that $x$ is conjugate to $y$. Thus $x \in \map {N_G} K$ and so $\order {\map {N_G} K} > 6$. As $6 \divides \order {\map {N_G} K}$ and $\order {\map {N_G} K} \divides 30$, it follows that: :$\map {N_G} K = G$ and so $K$ is normal in $G$. Thus: : $K$ and $\gen z$ are normal in $G$ : $K \cap \gen z = \set e$ : $K \gen z = G$ and it therefore follows from the  Internal Direct Product Theorem that: :$G = C_5 \times D_3$ {{qed}}