I am able to theoretically sink in the idea as given here, but still need few simple examples. I mean that examples be of real, integer, rational domain and binary operations defined on them; or of rotations. Need this to really be familiar with the intricacies involved with the idea.

  • $\begingroup$ Sorry but I am not completely sure: what are you exactly asking for? an example of "weird" group or an explanation of why a subgroup is defined how it is defined? $\endgroup$ – 57Jimmy Jan 24 '18 at 12:05
  • $\begingroup$ @57Jimmy I cannot understand why "'weird' group" is referred to by you. Need some more elaboration by what you meant. Do you mean that only 'weird' groups are possible in which a portion of the set of group is not a subgroup? $\endgroup$ – jiten Jan 24 '18 at 12:07
  • $\begingroup$ @57Jimmy I am clear about defn. of subgroups, but the closure property check is what confuses me. $\endgroup$ – jiten Jan 24 '18 at 12:16
  • $\begingroup$ you seem to be complaining, in your question, that the only examples you know are easy ones. That's why I've asked if you wanted some definition of more complicated ("weird") groups. $\endgroup$ – 57Jimmy Jan 24 '18 at 12:20
  • $\begingroup$ @57Jimmy I hope the edit makes the meaning clear, as I meant the reverse. $\endgroup$ – jiten Jan 24 '18 at 12:23

A group is just a bag of elements with an operation defined between them. (And a couple of special properties)

A subgroup is just a group inside a group. For example the integers with addition are a group. Take your bag full of integers and pick some integers, putting them in a second bag. Is that second bag a subgroup of the integers? Only if the second bag itself is a group! But for the second bag to be a group, the binary operation must make sense inside that bag.

I.e. take the second bag to be $\{1, 3, 4\} $. Does integer addition make sense in the universe of the second bag? Not really, because $3+4 = 7 \not\in \{1,3,4\} $

That is why you need a subgroup to be closed under the binary operation.

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    $\begingroup$ @jiten you can create more complex examples with a lot of ease. Just take groups you know and randomly pick a couple of elements. More often than not you won't get a group. Take $f(x) = x+2 $ and $g(x) = x^2$ from the group of real functions with function composition as binary operation. Is the set $\{f,g\} $ a subgroup? It doesn't even have the identity, but ignoring that fact, notice that $f(g (x)) = x^2+2 \not\in \{f,g \} $ $\endgroup$ – RGS Jan 24 '18 at 12:24
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    $\begingroup$ @jiten you should be able to modify my bag of functions such that it contains the identity and the inverses of the functions you put, and even so without the bag being closed under function composition. $\endgroup$ – RGS Jan 24 '18 at 12:42
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    $\begingroup$ You have $f(g(x)) = x+3$, which is neither the function $f$ nor the function $g$, so the set $\{f,g\}$ is not a subgroup, because it is not closed under function composition. (Also note that for function composition the identity is the function $i(x) = x$). $\endgroup$ – RGS Jan 24 '18 at 13:49
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    $\begingroup$ @jiten consider also $A:=\mathbb{Z} \backslash \{-1,1\}$. This has a $0$ and is closed under inversion, but it is not closed under the group operation, since for instance $3+(-2)=1$ and $1 \notin A$. $\endgroup$ – 57Jimmy Jan 24 '18 at 14:35
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    $\begingroup$ @jiten $f(x)$ has no identity of its own. $f$ is just an element in the bag of functions. In the bag of functions there is an element that is the identity for the operation, which is $i(x) = x$ and there is an element which is the inverse of $f$, namely $h(x) = x-2$: $f(h(x)) = h(f(x)) = x = i(x)$. $\endgroup$ – RGS Jan 24 '18 at 14:48

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