# Group theory with direct product

If $(G,*)$ and $(H,.)$ are groups, we can form a new group which is called the direct product $G \times H$ of $G$ and $H$, where the combination of two elements is defined by $(g_1,h_1)(g_2,h_2)=(g_1*g_2,h_1.h_2)$. Verify that $G \times H$ is a group.

I know the group axioms are closure, associativity, inverse and identity. But this way of presenting a group is new to me, and I can't use the usual approach.

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What is the usual approach? The most obvious approach I can think of for showing that something is a group is to go through the axioms and verifying them, and that works fine here. –  Tobias Kildetoft Jan 16 '13 at 19:42
It is pretty clearly verified in : en.wikipedia.org/wiki/Direct_product_of_groups#Definition –  MSEoris Jan 16 '13 at 19:42
what do you mean you can't use the usual approach? What is the usual approach? –  hmmmm Jan 16 '13 at 19:42
Well I'm more used to a group consisting of functions or matrices or defined by generators. How do I prove, for example, that G x H is associative? –  bbr4in Jan 16 '13 at 19:44

You rely almost entirely on the fact that $G, H$ are both groups, and their operations are thus associative, closed, each has an identity, and each element of the respective group has an inverse.

• Closure
Let $(g_1,h_1),(g_2,h_2)\in G\times H$ Then $(g_1,h_1)(g_2,h_2)=(g_1*g_2,h_1\cdot h_2)$ where $h_1\cdot h_2\in H$ and $g_1*g_2\in G$ so that, so $(g_1*g_2,h_1\cdot h_2) \in G\times H$

• Associativity:
Rely on the associativity of the respective operations of $G, H$. Let $g_1, g_2, g_3 \in G, h_1, h_2, h_3 \in H$ and show that $\Big((g_1,h_1)(g_2,h_2)\Big)(g_3,h_3)=(g_1, h_1)\Big((g_2, h_2)(g_3, h_3)\Big).$ Since the elements chosen are arbitrarily, this shows associativity holds over $G\times H$.

• Identity:
Consider $(e_G,e_H) \in G\times H$, $e_G \in G$ the identity of $G$, $e_H \in H$ the identity in $H$. Show for all $(g, h) \in G\times H, (g, h)(e_G, e_H) = (g, h)$, etc.

• Inverses:
Let $(g,h)\in G\times H$ so we have that $(g^{-1},h^{-1})\in G\times H$ since both $H$ and $G$ are groups and so for all $g \in G, g^{-1} \in G$, and likewise for $h \in H$, $h^{-1} \in H$. Then $(g, h)(g^{-1}, h^{-1}) = (g * g^{-1}, h \cdot h^{-1}) = (e_G, e_H)$, etc.

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What's the order of G x H? Is 'infinite' an acceptable answer? –  bbr4in Jan 16 '13 at 20:12
also, if G and H are both cyclic groups, is G x H also a cyclic group? –  bbr4in Jan 16 '13 at 20:14
The order of $G\times H = |G||H|$. So if G, H are both finite, so is $G\times H$. If one or both are infinite, then so is $G\times H.$ If both are cyclic, it is not necessarily the case that $G\times H$ is cyclic. $\mathbb{Z}_2 \times \mathbb{Z}_2$ is NOT cyclic, whereas $\mathbb{Z}_2$ is cyclic. The direct product of cyclic groups IS abelian. –  amWhy Jan 16 '13 at 20:14
Thank you. If P is a group of order 2, how many subgroups (trivial and proper) has the group P x P x P? –  bbr4in Jan 16 '13 at 20:34

To prove associativity, just do the computation:

\begin{align*} \Big((g_1,h_1)(g_2,h_2)\Big)(g_3,h_3)&=(g_1*g_2,h_1\cdot h_2)(g_3,h_3)\\ &=\Big((g_1*g_2)*g_3,(h_1\cdot h_2)\cdot h_3\Big)\\ &=\Big(g_1*(g_2*g_3),h_1\cdot(h_2\cdot h_3)\Big)\\ &=(g_1,h_1)(g_2*g_3,h_2\cdot h_3)\\ &=(g_1,h_1)\Big((g_2,h_2)(g_3,h_3)\Big)\;. \end{align*}

Each step is either from the definition of the group operation in $G\times H$ or from associativity of $*$ and $\cdot$ in their respective groups.

Everything else is an equally routine calculation using similar ideas.

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So we can start by veryfiying that the identity is in the group by considering the element $(e_G,e_H)$......

We can show closure by taking $(g_1,h_1),(g_2,h_2)\in G\times H$ and then we have $(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$ where $h_1h_2\in H$ and $g_1g_2\in G$ so that.....

We can show we have inverses by taking $(g_1,h_1)\in G\times H$ and then we have that $(g_1^{-1},h_1^{-1})\in G\times H$ as both $H$ and $G$ are groups and so have inverses.....

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seems the second element in your identity needs to be $\in H$ –  MSEoris Jan 16 '13 at 19:49
@MSEoris thanks, edited it now :) –  hmmmm Jan 16 '13 at 19:54
I thought the closure one is fine on its own, so there's no need for 'so that' perhaps? Same with the inverses. Can I consider $(1_G,1_H)$ for identity? –  bbr4in Jan 16 '13 at 19:55
@user52187 Well yes but you just need something along the lines of $(e_G,e_H)(g_1,h_1)=(e_Gg_1,e_Hh_1)=(g_1,h_1)$ –  hmmmm Jan 16 '13 at 20:01