# Statement(s) > [!NOTE] Theorem (Group Direct Product is a Group) > Let $(G,\circ_{1})$ and $(H,\circ_{2})$ be [[Groups|groups]]. Let $(G\times H,\circ)$ denote their direct product where $\circ$ is defined by $(g_{1},h_{1})\circ (g_{2},h_{2})= (g_{1}\circ_{1}g_{2},h_{1}\circ_{2}h_{2}).$Then $(G\times H,\circ)$ is also a group. # Proof(s) ###### Proof 1: (Associativity of Operation in Group Direct Product) Let $g_{1},g_{2},g_{3}\in G$ and $h_{1},h_{2},h_{3} \in H.$ Then $\begin{aligned}((g_1,h_1)\circ(g_2,h_2))\circ(g_3,h_3)&=((g_1\circ_1g_2)\circ_1g_3,(h_1\circ_2h_2)\circ_2h_3)&\text{Definition of $\circ$}\\&=(g_1\circ_1\left(g_2\circ_1g_3\right),h_1\circ_2\left(h_2\circ_2h_3\right))&\text{Associativity of $\circ_{1}$ and $\circ_{2}$}\\&=(g_1,h_1)\circ\left((g_2,h_2)\circ(g_3,h_3)\right)&\text{Definittion of $\circ$}\end{aligned}$ (Identity of Group Direct Product) Let $1_{G}$ and $1_{H}$ denote the identities of $G$ and $H$ respectively. Then $\begin{align} (1_{G},1_{H})\circ (g_{1},g_{2}) &=(1_{G}\circ_{1}g_{1} , 1_{H} \circ_{2} h_{1}) \\ &= (g_{1},h_{1}) \\ &= (g_{1}\circ_{1}1_{G}, h_{1} \circ 1_{H}) \\ &= (g_{1},h_{1})\circ(1_{G},1_{H}) \end{align}$and so $(1_{G},1_{H})$ is the identity element of $(G\times H,\circ).$ (Inverses in Group Direct Product) Clearly $(g_{1}^{-1},h_{1}^{-1})$ is the inverse of $(g_{1},h_{1}).$ # Applications