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.

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

I kind of have an idea that "having the Identity in the set" part of the definition of a group should be used to disprove it but I am having trouble formalising the proof. Any help would be really appreciated!


share|cite|improve this question
up vote 1 down vote accepted

A group needs to have a well-defined multiplication rule. In other words, you should be able to "multiply" any two elements and get another element of the group. In this case, the natural group operation would be given by composition, ie. you apply one transformation, then you apply the second, and the resulting transformation is the product.

What happens if you compose a reflection with itself? (ie, do the same reflection twice).

What happens if you compose a glide reflection with itself?

Edit: Now that the OP has figured it out, I should point out that this is what people refer to when they say a group should be "closed under its operation." I avoided this term, because I hate that this phrase is often included as a group axiom: a group operation is defined to be map $G\times G\to G$, so of course its image is in $G$. But here's an example where the operation takes you outside the underlying set, so the set is not closed under the operation.

share|cite|improve this answer
i guess you are saying that two reflections make a rotation/translation and since they are not in the group its not closed under composition. Is that what you are suggesting? – UH1 Dec 13 '12 at 3:31
Pretty much. I suggested composing a reflection with itself, which does nothing, rather than composing the reflection with another reflection, but it's the same problem: the product isn't an element of the set you're looking at. – Brett Frankel Dec 13 '12 at 3:33

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.