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January 9th, 2013, 04:17 PM  #1 
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Exercises about subgroups of an arbitrary group.
Hello to everybody, I'm a new member here, and I hope you can help me please with my exercises, since I'm not so good in math. I have the following exercise about subgroups of an arbitrary group: Let G be a group. Prove the following: Let H and K be subgroups of G. If H K, then H is a subgroup of K. I developed that on my own, in this way: By the definition of H K, we have that every element of H is also element of K. Since a subgroup is also a group, it is right to consider the following: the subgroup H of the group G, is subset of the group K. K is a subgroup of G, so the operation of K is the same of G. But we have that also H is a subgroup of G, so also the operation of H is the same of G. These elements imply that H,K have the same operation. we have also: i) H is nonempty subgroup of K, because he has at least the identity element which is common to K. ii) H is closed with respect to the same operation of K: , and since we have also that iii) H is closed with respect to the inverses: , and since we have also that so we have that H is subgroup of K. please can you tell me if it is correct? many thanks! 
January 9th, 2013, 09:22 PM  #2 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Re: Exercises about subgroups of an arbitrary group.
you don't need to show closure with respect to K, existence of an identity and inverses. you are given already that: H is a subgroup of G. this means H is a group in its own right (its operation is just the operation of G, restricted to H), so because it's a group, already has ALL the properties of a group, including: a) closure on * (the operation of G) on H, which is associative since its associative for ALL elements in G (including those that happen to be in H). b) existence of an identity (which is the same one as G has). c) existence of inverses in H (which are the same inverses as they are in G). since K also has the same operation as G (and thus the same operation as H), you obtain H as a subgroup of K, by first restricting * to K, and then further restricting it to H (which is kosher, since H is a subset of K). H is a subgroup of ANY subgroup of G that contains it. in fact, we can go the other way: if H is a subgroup of K, and K is a subgroup of G, then H is a subgroup of G. 
January 10th, 2013, 02:20 AM  #3  
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Re: Exercises about subgroups of an arbitrary group.
thanks! Quote:
Quote:
H is a group in its own right (its operation is just the operation of K, restricted to H), so because it's a group, already has ALL the properties of a group, including: a) closure on * (the operation of K) on H, which is associative since its associative for ALL elements in K (including those that happen to be in H) since H subset of K. b) existence of an identity (which is the same one as K has) since H subset of K. c) existence of inverses in H (which are the same inverses as they are in K) since H subset of K. is it right?  
January 10th, 2013, 10:58 AM  #4 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Re: Exercises about subgroups of an arbitrary group.
not quite. you must also USE the fact that: H is a subset of K. K is a subgroup of G (given). thus the operation of K is the operation of G, restricted to K. now H is a subSET of K. but H is a subGROUP of G. so the whole proof is: * (restricted from G to H) = * (restricted from G to K, and then restricted from K to H). to understand better what i mean, let's use functional notation instead of "infix" notation. we have a function m:GxG>G, given by m(a,b) = a*b since K is a subgroup of G: the function m':KxK>G given by m'(a,b) = a*b is closed, that is: im(m') ? K. now consider the function m":HxH>G given by m"(a,b) = a*b. since H is a subset of K, and since m"(a,b) = m'(a,b) for all a,b in H, we know that im(m") ? K. but since H is a subgroup of G, the function m":HxH>G actually has im(m") ? H. we have two paths that leads to the same result: m>m" (restricting from G to H) m>m'>m" (restricting first from G to K, then from K to H) the reason we can conclude that im(m") ? H, is that for H to be a group, m" must be a function from HxH>H (definition of a binary operation). ************* all of this makes it sound much more complicated then it really is. let me give an example: suppose G = Z, the group of integers under addition. let H = 4Z, all multiples of 4, and K = 2Z, all multiples of 2 (even numbers). note that every multiple of 4 is even, so 4Z ? 2Z. we thus conclude that the multiples of 4 are not just a subset of the even numbers, but actually a subgroup. the important part of the phrase "X is a subgroup of Y" is not "Y" but "subgroup". 
January 11th, 2013, 02:04 AM  #5  
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Re: Exercises about subgroups of an arbitrary group. Quote:
Quote:
Quote:
or maybe, have I misundersood this point?  
January 11th, 2013, 10:22 AM  #6 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Re: Exercises about subgroups of an arbitrary group.
one of the ways we check that a subset H of a group G is a subgroup is that * is closed on H. because a group must have a binary operation *:GxG>G so if a subset of G is a group, with respect to the operation * of G,then * when restricted to H, must induce a binary operation on H, that is *HxH>H. this is the same thing as closure. that is: saying (H,*) is a subgroup of (G,*) is the same thing as saying: H is a subset of G, and (H,*) is a group. it IS important that the operation be "the same one". for example {1,1} is certainly a subset of Z, but is not a subgroup of (Z,+), although it IS a group under multiplication. 
January 12th, 2013, 03:28 AM  #7  
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Re: Exercises about subgroups of an arbitrary group.
ok it's clear. Quote:
and also if i understand correcly, the two path that you are saying, are not two alternative options (as I comprehended previously) but one way with one path followed by the other one. In fact briefly: >Since G is a group we must have an binary operation *:GxG>G >Since K is a subgroup of G, the operation * of K is the same of G, and since K is a subgroup of G, it is closed respect with the * of G. >Since H is a subgroup of G, the operation * of H is the same of G, and since H is a subgroup of G, it is closed respect with the * of G. >H and K has the same * of G. >the operation * of K is the same of G, and H it's closed with respect the * of G (being H subgroup of G, it is also a group and then the result of the operation must reside in H itself) , and since H subset of K the result of the operation on H is in K. So H is closed with respect the operation * of K. Quote:
or not?  
January 14th, 2013, 10:22 AM  #8 
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  Re: Exercises about subgroups of an arbitrary group.
ok I've resolved. Many thank you very much for your explanation! 

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