Proof in a topological space

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• Jan 22nd 2006, 02:48 AM
rgep
Quote:

Originally Posted by ThePerfectHacker
... the union/intersection of two subgroups is a subgroup.

Not so: the union of two subgroups is almost never a subgroup (exercise: when exactly?).
• Jan 22nd 2006, 03:50 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
CaptainBlack what you said is true. I made a mistake in how I posed my question, I wanted to say a subgroup not a subset.

The revised question:
1)Let $\displaystyle G$ be a group.
2)Let $\displaystyle P(G)=\{X|X\leq G\}$.
3)Let $\displaystyle S=\{\} \cup P(G)$.
4)Then $\displaystyle S$ is a topology on $\displaystyle G$.
The reason is because the union/intersection of two subgroups is a subgroup.

You had better tell us what you want the expressions "union of two subgroups"
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 AM
ThePerfectHacker
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 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
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.

But $\displaystyle S'$ is not in general closed under the group operation, and
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 PM
ThePerfectHacker
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 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
By the way, I like your group multiplication box, that probably took a long time in Latex code.

I like it too, but it didn't take long at all, the examples in the manual are
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 PM
ThePerfectHacker
But it is true that the intersection of subgroups is a subgroup.
• Jan 24th 2006, 07:50 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
But it is true that the intersection of subgroups is a subgroup.

Apparnetly.

RonL
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