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I've been studying about subgroups and I encountered an example with answers and does not have explanation how it is derived and I need help to understand it.

Here is the example:

Example 1.4.20

Determine whether the given subset of the complex numbers is a subgroup of the group $\mathbb{C}$ of complex numbers under addition.

a.) $\mathbb{R}$: YES

b.) $\mathbb{Q}^+$: NO, there is no identity element.

c.) $7\mathbb{Z}$: YES

d.) The set of $i\mathbb{R}$ of pure imaginary numbers including $0$: YES

e.) The set $\pi\mathbb{Q}$ of rational multiples of $\pi$: YES

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    $\begingroup$ Are there particular letters which are confusing you, or are all of the examples equally mysterious? $\endgroup$ – Trold Oct 9 '14 at 2:11
  • $\begingroup$ I am new to this and our professor is so lazy in explaining and just sits on his chair without chalk writing in the board and we are all confused. That's why I have been studying and the book also is so confusing. I am confused why it can say that a.) Yes, b.) No c.) Yes d.) Yes and e.) No without support that can make me understand. I am really sorry if I am so dumb with this. It really makes me crazy. $\endgroup$ – MODULUS Oct 9 '14 at 2:16
  • $\begingroup$ I just want to know like this : Example 10(7+3)=100 because 10(10) is 100. I need to know a supporting detail of the answer why it is like that. $\endgroup$ – MODULUS Oct 9 '14 at 2:18
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For a subset to be a subgroup, it has to be closed under the group's binary operation and the formation of inverse. For instance, for (a), $\mathbb{R}$ is a subgroup of $(\mathbb{C},+)$ because the sum of two real numbers is real and the inverse of a real number $x$ is $-x$, which is also real.

For (b), the answer is no because the identity element of the group is $0$, and it does not belong to $\mathbb{Q}^+$. Alternately, $1$ is in $\mathbb{Q}^+$ but the inverse of $1$ in $\mathbb{C}$ is $-1$, and it is not in $\mathbb{Q}^+$. So $\mathbb{Q}^+$ is not closed under taking of inverse, and is not a subgroup.

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  • $\begingroup$ In (b) the identity element is 0 why in the given example it says "there is no identity element" ? $\endgroup$ – MODULUS Oct 9 '14 at 2:29
  • $\begingroup$ This is because $0$ is not in $\mathbb{Q}^+$, so $\mathbb{Q}^+$ has no identity element. $\endgroup$ – E W H Lee Oct 9 '14 at 2:30
  • $\begingroup$ Oh, I thought when it says no identity element it means no at all. Thank you. Can you also please help me understand on d.) and e.? $\endgroup$ – MODULUS Oct 9 '14 at 2:38
  • $\begingroup$ Note that $i\mathbb{R} = \{ ia | a \in \mathbb{R}\}$. Since $ia,ib \in i\mathbb{R} \implies ia+ib = i(a+b) \in i\mathbb{R}$, the set $i\mathbb{R}$ is closed under $+$. Also, it is clear that $ia \in i\mathbb{R} \implies -ia = i(-a) \in i\mathbb{R}$. So $i\mathbb{R}$ is closed under taking inverses. Hence $i\mathbb{R}$ is a subgroup of $(\mathbb{C},+)$. Similar arguments apply to (e). $\endgroup$ – E W H Lee Oct 9 '14 at 2:42
  • $\begingroup$ Thank you for helping. Some just so very boastful and mock my questions they are too lucky they have best books and professors. $\endgroup$ – MODULUS Oct 9 '14 at 2:53
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Here's a (maybe too verbose) explanation:

A subgroup is a special subset of a group, specifically it's special because it forms a group in its own right (under the same operation as the group containing it).

Example: We know, or can quickly check that $\mathbb{C}$ (the complex numbers) is a group under addition. I'm not sure what you're Prof.'s favorite version of the group axioms are, but here's one version of them:

  • The operation $+$ is associative: $a+(b+c)=(a+b)+c$ i.e., it doesn't matter if we add $b$ and $c$ first or $a$ and $b$ first.
  • $\mathbb{C}$ is closed under addition, adding any two complex numbers is going to give you another complex number
  • The number $0$ acts as the identity element ($0+x=x$ for any $x\in\mathbb{C}$).
  • Every number $x$ has an additive inverse, namely $-x$.

(Note that this list isn't as short as it could be, but I think it's about right for someone just learning groups)

We know that the reals are contained in $\mathbb{C}$, so $\mathbb{R}$ is a subset of $\mathbb{C}$, but it's a subset which also satisfies these four axioms of its own.

  • Addition is associative comes free from the fact that it's associative in $\mathbb{C}\supset\mathbb{R}$
  • $0$ is a real number, so the identity is inside $\mathbb{R}$.
  • If you add two reals, you get a real so closure under the operation holds, and
  • If you negate an element of $\mathbb{R}$, you get another element of $\mathbb{R}$, so every inverse of an element of $\mathbb{R}$ is also in $\mathbb{R}$.

By contrast, the group of positive rational numbers $\mathbb{Q}^+$ is not a subgroup of $\mathbb{C}$ because it does not contain the identity element. ($0$ isn't positive)

Hopefully that's a good start, if there are other examples on the list that might help, please say so in comments.

Edit in response to comment:

d) and e) are almost identical in what their justifications look like, so I'll put up d) and leave e) to you.

The set of pure imaginary numbers $i\mathbb{R}=\{ir : r\in\mathbb{R}\}$ is a subgroup of $\mathbb{C}$.

  • Associativity (as always) comes as a freebie from the associativity of $+$ on $\mathbb{C}$. (Honestly, you can probably after a while drop this from subgroup questions because it never fails)
  • The pure imaginary numbers are closed under addition. Suppose we have $ir_1$, $ir_2\in i\mathbb{R}$. Then $ir_1+ir_2=i(r_1+r_2)$. Since $r_1+r_2\in\mathbb{R}$, $i(r_1+r_2)\in i\mathbb{R}$.
  • The inverse of a pure imaginary number is also a pure imaginary number. Take $ir$ for some $r\in\mathbb{R}$. Then $-r\in\mathbb{R}$ and $-ir\in i\mathbb{R}$.
  • We're given that the identity is in $i\mathbb{R}$, but it probably wouldn't hurt for you to point out why.
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  • $\begingroup$ The d.) and e.) scares me. Can you also please explain it with steps and supporting theorems. You are right your axioms is so clear. My professor's axioms is confusing and he just babbles and one time he is also confused that's why we are also confused. $\endgroup$ – MODULUS Oct 9 '14 at 2:35
  • $\begingroup$ @Trigo I've edited my post to include a work-through of (most of) example d). $\endgroup$ – Trold Oct 9 '14 at 2:45

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