0
$\begingroup$

$i$ is the imaginary unit on the complex plane.

I am confused on a particular portion of the definition of a subgroup.

Let $G$ be a group and $H$ is a subset of $G$.

A part of the definition is stated as follows: "The restriction of the group operation to $H$ takes values in $H$"

Let $+: G \times G \to G$

So I am interpreting that portion of the definition as $+|_H: H \times H \to G$ or is the codomain supposed to $H$? That's what I'm not clear on...

And how $\mathbb{Q}[i]$ is not a subgroup of reals... Thanks! I just started to learn on group theory so details would be appreciated, otherwise i would not understand...

Also from the "one step test of subgroups" I don't think it fails?

$\endgroup$
1
  • 4
    $\begingroup$ Because $i\notin \mathbb{R}$ $\endgroup$
    – saulspatz
    Commented May 18, 2019 at 0:01

1 Answer 1

Reset to default
4
$\begingroup$

$\mathbb{Q}(i)$ is not a subsets of the set of real numbers, so it is not a subgroup.

$\endgroup$
2
  • $\begingroup$ OHHH, it fails the condition of the subgroup being a subset of the group correct? And is my interpretation of the definition correct? $\endgroup$
    – javacoder
    Commented May 18, 2019 at 0:02
  • 2
    $\begingroup$ your interpretation is correct $\endgroup$
    – Tsemo Aristide
    Commented May 18, 2019 at 0:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .