My Math Forum Exercises about subgroups of an arbitrary group.

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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 $\subseteq$ K, then H is a subgroup of K. I developed that on my own, in this way: By the definition of H $\subseteq$ 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 non-empty 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: $ab \in H$, and since$H \subseteq K$ we have also that $ab \in K$ iii) H is closed with respect to the inverses: $a,a^{-1} \in H$, and since$H \subseteq K$ we have also that $a,a^{-1} \in K$ 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
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Re: Exercises about subgroups of an arbitrary group.

thanks!
Quote:
 Originally Posted by Deveno you don't need to show closure with respect to K, existence of an identity and inverses
so it is useless for the purpose of the exercise?

Quote:
 Originally Posted by Deveno 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).
so is it correct to say that in these terms (using words as seen previous)?:
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
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Re: Exercises about subgroups of an arbitrary group.

Quote:
 Originally Posted by Deveno 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)
if I have understood the second path is the following that you have described previously. Where is the first path?

Quote:
 Originally Posted by Deveno 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.
Quote:
 Originally Posted by Deveno 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).
maybe, is the correct word subgroup instead of group, because of his closure respect with the operation * ?
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
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Re: Exercises about subgroups of an arbitrary group.

ok it's clear.
Quote:
 Originally Posted by Deveno 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).
I was translating in a wrong way, sorry... I can say also that: "since H is also a group". In this phrase you are underlining the fact that a subgroup is also a group.

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:
 Originally Posted by Deveno 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".
so referring to the group K. it is important that the operation of H be the same one of K. As I said above.

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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