Not so: the union of two subgroups is almost never a subgroup (exercise: when exactly?).Quote:

Originally Posted byThePerfectHacker

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- Jan 22nd 2006, 02:48 AMrgepQuote:

Originally Posted by**ThePerfectHacker**

- Jan 22nd 2006, 03:50 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

and "intersection of two subgroups" to mean. As it is they don't mean

anything as union and intersection are usually defined on sets. And a

groups is not a set, a set may be involved but the group is not the set,

and as rgep points out the naive union/intersection of the associated sets

of a pair of subgroups is not usually the set associated with another subgroup.

RonL - Jan 22nd 2006, 10:12 AMThePerfectHacker
Maybe I did make a mistake. I just remember my book on Abstract Algebra asking to prove the intersection/uninon of groups, is a group something like that. Maybe, it is true for

**Normal Subgroups**I do not remember exactly.

I do not think I made a mistake. By uninon/intersection I mean $\displaystyle S\cup S'=\{x|x\in S \mbox{ or/and } x\in S'\}$ respectively. I just am visualizing that in my head, I think it is a subgroup. - Jan 22nd 2006, 10:58 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

so in general not a subgroup.

You might want to consider the group $\displaystyle \mathcal{G}$ with the following group multiplication

table (which if I have done this right makes $\displaystyle \mathcal{G}=(\{I,a,b,c\},\otimes)$ a group):

$\displaystyle \begin{array}{c|ccccc}\otimes&I&a&b&c\\ \hline I&I&a&b&c\\a&a&I&c&b\\b&b&c&I&a\\c&c&b&a&I\end{arr ay}$

Then: $\displaystyle (\{I,a\},\otimes)$, $\displaystyle (\{I,b\},\otimes)$, $\displaystyle (\{I,c\},\otimes)$, are all subgroups of $\displaystyle \mathcal{G}$,

but $\displaystyle (\{I,a\} \cup \{I,b\},\otimes)$ is not a group because $\displaystyle \{I,a\} \cup \{I,b\}$

is not closed under $\displaystyle \otimes$

RonL - Jan 22nd 2006, 01:12 PMThePerfectHacker
Thank you CaptainBlack, my only mistake was that I assumed that addition was closed.

Okay now, change the problem, which type of subgroups will form a subgroup?

By the way, I like your group multiplication box, that probably took a long time in Latex code. - Jan 22nd 2006, 01:28 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

easy to follow, which is what I did once I had worked out the group table

that I wanted.

RonL - Jan 22nd 2006, 02:17 PMThePerfectHacker
But it is true that the intersection of subgroups is a subgroup.

- Jan 24th 2006, 07:50 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

RonL