If ab is in a subgroup, are a and b neccessarily in the subgroup?

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Homework Help Overview

The discussion revolves around the properties of subgroups in group theory, specifically questioning whether the product of two elements in a subgroup implies that both elements are also in the subgroup.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the implications of the subgroup property regarding the elements a and b when their product ab is known to be in the subgroup. Some participants question the validity of the reverse implication and discuss related properties of group elements.

Discussion Status

Participants are exploring the implications of subgroup properties, with some providing insights into the one-way nature of the relationship. There is an ongoing exchange of ideas regarding the inverse of products and their relevance to the original question.

Contextual Notes

Participants are considering the definitions and properties of groups and subgroups, including closure under multiplication and the identity element, while acknowledging the limitations of these properties in the context of the original question.

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Homework Statement


I was just wondering if the product ab is in a subgroup, are a and b necessarily in the subgroup, as well?

The Attempt at a Solution


I think they are, but how would you prove that? Or is that obvious from closure under multiplication and you don't need to prove it? I know it works the "normal" way, that is if a and b are in the subgroup (or group), then ab is in it, as well, but I'm not sure about the reverse direction.

Thanks in advance for any replies.
 
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The identity e is in every subgroup. a*a^(-1)=e for all elements of the group. That doesn't mean a is in every subgroup.
 
Ah, I see, so it only works one way then? Thanks for the quick response.
 
Oh, sorry, one more thing, (ab)-1 is always b-1a-1, though, right?
 
Ryker said:
Oh, sorry, one more thing, (ab)-1 is always b-1a-1, though, right?

Sure. Multiply b^(-1)a^(-1) and ab. What do you get?
 
Yeah, that's what I was thinking :smile:
 

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