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I am a little confused in the interpretation of the product of groups. Here is what's written in my notes:


Given groups $G_1,G_2,...,G_n$, recall that $G_1\times G_2\times ...\times G_n=\{(g_1,g_2,...,g_n)\}\vert g_i \in G_i \}$ for all $i \in I$. More generally, for groups $G_i$ indexed $i \in I$, we have $\Pi_{i\in I}G_i=\{(g_i)_{i\in I}\vert g_i \in G_i$ for all $i\}$.


So if I let $G_1=(*,X)$ where $X=\{something\}$ and $G_2=(*,Y)$ where $Y=\{something\}$, according to the definition above, what would $G_1 \times G_2 $ yield?

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  • $\begingroup$ Your last $X$ was meant to be $Y$? $\endgroup$ – Gerry Myerson May 13 '13 at 12:40
  • $\begingroup$ @GerryMyerson. Yes, sorry. Fixed. $\endgroup$ – Gustavo Louis G. Montańo May 13 '13 at 12:42
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If $G_{1}=(X,\ \ast)$ and $G_{2}=(Y,\square)$ are groups then $((G_{1}\times G_{2}, \cdot)$ is the group given by the set $X\times Y=\{(x,y):\ x\in X\ \land\ y\in Y\}$ and the operation $\cdot$ on it defined as $$\forall (x,y),\ (w,z)\in X\times Y,\ (x,y)\cdot (w,z):=(x\ast w,\ y\square z).$$ If $G_{1}=G_{2}=(X, \ast)$ this just becomes the set $X\times X$ with the operation $(x,x_{1})\cdot (x',x_{1}')=(x\ast x', x_{1}\ast x_{1}')$ for all $(x,x_{1}),\ (x',x_{1}')\in X\times X$.

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  • $\begingroup$ sorry man, I haven't seen you were writing $\endgroup$ – Edoardo Lanari May 13 '13 at 12:48
  • $\begingroup$ Same thing for me. $\endgroup$ – Marco Vergura May 13 '13 at 13:42
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The canonical construction of the product of two groups yields the group $ (G_{1} \times G_{2},*,1_{G_{1} \times G_{2}},^{-1})$ where the underlying set is the cartesian product $G_{1} \times G_{2}$ and the binary operations is defined by $(g_{1},g_{2})*(h_{1},h_{2}):=(h_{1}*g_{1},h_{2}*g_{2})$ (I use the same symbol for those operations but their meaning should be clear from the context). Moreover $(g_{1},g_{2})^{-1}=(g_{1}^{-1},g_{2}^{-1})$.

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