$\{G,*\}$ is a group with identity element $e$, and $\{G',\circ\}$ is a group with identity element $e'$. Let $S=G\times G'$. Define the “product” of pairs of elements $(a,a'),(b,b')\in S$ by $(a,a')(b,b')=(a\circ b,a'*b')$

Prove that $S$ is a group under the “product” operation.

My first thoughts on the problem was that how can we prove that $a\circ b\in G$ and $a'*b'\in G'$ and thus that $S$ is closed under the "product" operation. The problem is, I can't seem to find a way to do so. The fact that $a*b\in G,\forall a,b\in G$ doesn't seem to give any information about whether $a\circ b\in G,\forall a,b\in G$. Any help with regards to closure would be helpful. Hopefully I can use that help to figure out associativity, identity and inverse for myself.

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    $\begingroup$ Surely the question as posted contains a typo, as $\circ $ is not even defined on $G.$ $\endgroup$ – user17794 Aug 6 '12 at 3:49
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    $\begingroup$ @TimDuff: I think that's what OP is confused about. $\endgroup$ – Kris Aug 6 '12 at 3:53

I believe you need only show that for $a,b \in G, \, c,d \in G' \Rightarrow (a*b, c\circ d) \in G \times G'$.

  • $\begingroup$ You got me by a minute William $\endgroup$ – Moderat Aug 6 '12 at 3:54
  • $\begingroup$ I think you got the product operation written down incorrectly. It is supposed to be $(a,a')(b,b')=(a\circ b,a'*b')$ not $(a,a')(b,b')=(a*b,a'\circ b')$ $\endgroup$ – E.O. Aug 6 '12 at 3:59
  • $\begingroup$ @E.O. $\circ$ is not an operation on $G$ so what you wrote doesn't make sense. $\endgroup$ – Ink Aug 6 '12 at 4:23
  • $\begingroup$ @E.O. the OP wrote the groups as $(G,*)$ and $(G', \circ)$ therefore my answer is correctly written. Regards. $\endgroup$ – Moderat Aug 6 '12 at 4:25
  • $\begingroup$ @Brian that is what I thought too, but since the book I am using seems to have 2 other questions based on the same binary operator I was not sure. $\endgroup$ – E.O. Aug 6 '12 at 4:38

I switch the group operation symbol. I used $(G, \circ)$ and $(G', *)$.

Let $\cdot$ be the product operation on $G \times G'$. That is $(a,b)\cdot (a',b') = (a \circ a', b * b')$. Clearly $G \times G'$ is closed under the $\cdot$ since $a \circ a' \in G$ and $b * b' \in G'$.

$\cdot$ is associative. Suppose $a, a', a'' \in G$ and $b,b', b'' \in G'$. Then

$(a,b) \cdot ((a',b') \cdot (a'', b'')) = (a,b) \cdot (a' \circ a'', b' * b'') = (a \circ (a' \circ a''), b * (b' * b''))$

using the associativity of $\circ$ and $*$, you have

$= ((a \circ a') \circ a'', (b * b') * b'') = (a \circ a', b * b') \cdot (a'', b'') = ((a,b) \cdot (a',b')) \cdot (a'', b'')$

Thus associativity has been shown.

Similarly using individual operations you can show that $(e, e')$ is the identity of the product group. I leave it you to figure out what the inverse would be.

  • $\begingroup$ Are you sure you are not making a mistake when you say that $a \circ a' \in G$? I think that it should be $a \circ b \in G$ under your (switched) notation. $\endgroup$ – Moderat Aug 6 '12 at 5:10
  • $\begingroup$ @jmi4 I believe this typo should only affect the second paragraph. It is fixed now. $\endgroup$ – William Aug 6 '12 at 5:39

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