Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 : – 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
up vote 5 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

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.....

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.