Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\{G,*\}$ 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.

share|improve this question
1  
Surely the question as posted contains a typo, as $\circ $ is not even defined on $G.$ –  user17794 Aug 6 '12 at 3:49
1  
@TimDuff: I think that's what OP is confused about. –  Kris Aug 6 '12 at 3:53

2 Answers 2

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.

share|improve this answer
    
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. –  Zvpunry Aug 6 '12 at 5:10
    
@jmi4 I believe this typo should only affect the second paragraph. It is fixed now. –  William Aug 6 '12 at 5:39

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

share|improve this answer
    
You got me by a minute William –  Zvpunry Aug 6 '12 at 3:54
    
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 $(a,a')(b,b')=(a*b,a'\circ b')$ –  E.O. Aug 6 '12 at 3:59
    
@E.O. $\circ$ is not an operation on $G$ so what you wrote doesn't make sense. –  Vectk Aug 6 '12 at 4:23
    
@E.O. the OP wrote the groups as $(G,*)$ and $(G', \circ)$ therefore my answer is correctly written. Regards. –  Zvpunry Aug 6 '12 at 4:25
    
@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. –  E.O. Aug 6 '12 at 4:38

Your Answer

 
discard

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.