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NotNotBatman
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Prove that if  N \leq G and  |G:N| = 2 then N \trianglelefteq G .

For a proof where I don't know where to start, I'd list properties derived from info given in the question, that is

N subgroup of G
 e_G \in N
 \forall n \in N, \exists n^{-1} \in N, s.t. , n^{-1}n = e
 \forall n_1, n_2 \in N, n_1n_2 \in N

Index N in G = 2

 G = g_1N \cup g_2N for some  g_1,g_2 \in G
From this we know  g_1N \cap g_2N = \varnothing


and we want to show  gN=Ng, \forall g \in G.

Here's the thing with these proofs, unless it's an obvious one like showing equality, bijection etc. I never know what to do and I can try so many things but I never seem to get to the right answer. I've looked at the first line of the answer and it says let  g \in G \setminus N .
But I have no idea how to decide this and the answer never seems to follow from the things we can deduce as I've written above.

So in addition to the proof above; how does the thinking process for these proofs work? Because my way of doing it hasn't worked for the past 2 years.
Last edited by NotNotBatman; 4 weeks ago
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zetamcfc
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(Original post by NotNotBatman)
Prove that if  N \leq G and  |G:N| = 2 then N \trianglelefteq G .

For a proof where I don't know where to start, I'd list properties derived from info given in the question, that is

N subgroup of G
 e_G \in N
 \forall n \in N, \exists n^{-1} \in N, s.t. , n^{-1}n = e
 \forall n_1, n_2 \in N, n_1n_2 \in N

Index N in G = 2

 G = g_1N \cup g_2N for some  g_1,g_2 \in G
From this we know  g_1N \cap g_2N = \varnothing


and we want to show  gN=Ng, \forall g \in G.

Here's the thing with these proofs, unless it's an obvious one like showing equality, bijection etc. I never know what to do and I can try so many things but I never seem to get to the right answer. I've looked at the first line of the answer and it says let  g \in G \setminus N .
But I have no idea how to decide this and the answer never seems to follow from the things we can deduce as I've written above.

So in addition to the proof above; how does the thinking process for these proofs work? Because my way of doing it hasn't worked for the past 2 years.
Think about the co-sets you will have. One will be N itself (For both left and right), what is the other (For both left and right). If we take an element of G what co-sets can it form with N?

The thinking for proofs like these is quite simple, we just take the definitions we know along with the supposition and fairly quickly we arrive at the conclusion. For things like this there is no trick you nee to see, so you can stick to the main road and not overcomplicate things. But if you are stuck try to think about starting at the conclusion you want and working backwards seen as it is a problem you know has been solved so going back in logical stages is fine, not so great if you have no idea whether or not a statement is true or false.
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NotNotBatman
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(Original post by zetamcfc)
Think about the co-sets you will have. One will be N itself (For both left and right), what is the other (For both left and right). If we take an element of G what co-sets can it form with N?

The thinking for proofs like these is quite simple, we just take the definitions we know along with the supposition and fairly quickly we arrive at the conclusion. For things like this there is no trick you nee to see, so you can stick to the main road and not overcomplicate things. But if you are stuck try to think about starting at the conclusion you want and working backwards seen as it is a problem you know has been solved so going back in logical stages is fine, not so great if you have no idea whether or not a statement is true or false.

After reading this, what I'm thinking now is that to split G into two groups, we can consider the disjoint union  G = N \cup G \setminus N , so we can form the coset N and now we need to form the coset  G \setminus N using some coset of N in G. Not sure how to do this and how to show N normal to G. Even going backwards in logical steps doesn't help me.
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zetamcfc
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(Original post by NotNotBatman)
After reading this, what I'm thinking now is that to split G into two groups, we can consider the disjoint union  G = N \cup G \setminus N , so we can form the coset N and now we need to form the coset  G \setminus N using some coset of N in G. Not sure how to do this and how to show N normal to G. Even going backwards in logical steps doesn't help me.
Well, let us take a general g in G. If g is in N what must gN an Ng be? If g is not in N what must gN an Ng be?
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NotNotBatman
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(Original post by zetamcfc)
Well, let us take a general g in G. If g is in N what must gN an Ng be? If g is not in N what must gN an Ng be?
Well, if g is in N then all elements in gN and Ng are in N.

For the latter, I'm not sure, because how do I know that G\N is a subgroup? So I dont know if I can apply the closure axiom.
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zetamcfc
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(Original post by NotNotBatman)
Well, if g is in N then all elements in gN and Ng are in N.

For the latter, I'm not sure, because how do I know that G\N is a subgroup? So I dont know if I can apply the closure axiom.
We know the only co-sets are N an G\N. If g is not in N then gN an Ng cannot be N therefore they must be...
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NotNotBatman
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(Original post by zetamcfc)
We know the only co-sets are N an G\N. If g is not in N then gN an Ng cannot be N therefore they must be...
Oh. They must be G\N
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