# Why is this sufficient proof that T is a subgroup? [duplicate]

Exercise: Let $$C∗$$ be the multiplicative group of complex numbers and suppose we let $$T = \{z \in C~\big|~ |z|^2 = 1\}$$. Show that $$T$$ is a subgroup of $$C∗$$ (i.e. for any $$z = a * bi, a^2 + b^2 = 1$$).

In my notes it says that we can easily prove this by showing $$x, y \in T$$ and show that $$x*\text{inverse}(y) ∈ T$$. How does this imply that $$T$$ satisfies the properties required of a group (inverse/identity, associativity)?

• $xx^{-1}=e$, $ex^{-1}=x^{-1}$, $x(y^{-1})^{-1}=xy$. Also, subgroup means closed under identity, inverse, and the operation; associativity is a given.
– anon
Commented Oct 16, 2022 at 3:20
• This is called the subgroup test, and the answer to your question is, essentially, the proof of this face Commented Oct 16, 2022 at 3:23
• Hello Lillian, welcome to stackexchange. Commented Oct 16, 2022 at 3:26
• Likely duplicate question math.stackexchange.com/questions/2425905/… Commented Oct 16, 2022 at 3:32

The seems a really weird way of putting it but it would work.

If for all $$x,y \in T$$ we always have $$xy^{-1}\in T$$ then

For any $$x,x\in T$$ then $$xx^{-1} = e\in T$$ and therefore for any $$x\in T$$ then $$e\cdot x^{-1} =x^{-1} \in T$$.

And finally if $$x,y\in T$$ then $$y^{-1} \in T$$ (as we showed above) so $$x(y^{-1})^{-1} = xy\in T$$.

That's all we need.

We've proven $$\cdot$$ is closed (but IMO in a really round about way). Associativity is inherited. The identity element of $$\mathbb C$$ is the identity element of $$T$$ and we've proven (very round about way) that $$e=1\in T$$. And we have proven for every $$x\in T$$ that $$x^{-1}\in T$$.

That's all that is needed to show its a group.

.....

But in my opinion this "let's do it by proving only one thing" is cutting your nose of to spite your face. And I'm not sure how you'd prove $$xy^{-1}\in T$$ without first showing $$y^{-1}\in T$$ and frankly proving the TWO items 1) $$|1|=1$$ so $$1 \in T$$ and 2) if $$|x|=1$$ then $$|x^{-1}| = 1$$ so for all $$x \in T$$ we know $$x^{-1}\in T$$, would be a proof that is easier, more direct, and much more readable.

We already have that the identity is in the group as $$1$$, and we already have that the inverse is in the subgroup, since it's clear that if $$|z|=1$$ then $$|\bar z|=1$$, and $$\bar z=z^{-1}$$, and since these are complex numbers the associativity part is also satisfied automatically. So it only remains to demonstrate closure.

So while you are right that all of these need to be demonstrated, the professor has assumed that you would notice they are all clearly true without explicitly writing them out.

Incidentally these numbers are all of the form $$\cos\theta+i\sin\theta$$ of course.

• While true, it's better to say that $H\leq G$ if and only if for every $x,y\in H$, we have $xy^{-1}\in H$. This is the general criterion for subgroup, and there is nothing special going on here to show $S^1 \leq \mathbb C^*$. Commented Oct 16, 2022 at 3:52
• @AndrewZhang I was going to delete this answer but I've kept it here since you cannot see deleted posts and I think we should keep this since your comment is more informative than the upvoted answer. Commented Oct 16, 2022 at 4:47

Proofs vary in how much thoroughness is required by your professor, course, etc.

The following two steps are sufficient for proving a subgroup is a group

• Show the subgroup is closed

• Show the inverse for each element is in the subgroup.

What is often assumed in these proofs:

• Associativity is inherited from the group.
• The basic properties of the operation are inherited from the group.
• Closure in the subgroup and inverse in the subgroup each must be proven. Together, these imply $$a * a^{-1}$$ = 1, therefore the identity is in the subgroup.
• Because the subgroup contains the identity, or pick your element, it is nonempty.