Let $\mathbb{ℂ^∗}$ = $\mathbb{ℂ}$\ ${0}$ denote the group of non-zero complex numbers under multiplication. Suppose
$Y_n$={$z\in \mathbb{C} |z^n=1$}
then which of the following is (are) subgroups of $\mathbb{C^*}$
1).$\bigcup_{n=1}^{100}Y_n$
2).$\bigcup_{n=1}^{\infty}Y_{2^n}$
3).$\bigcup_{n=100}^{\infty}Y_{n}$
4).$\bigcup_{n=1}^{\infty}Y_{n}$
solution i tried - Here $Y_{n}$ is a cyclic subgroup of $\mathbb{C^*}$ but as we know that union of two subgroup is a subgroup iff one is contained in other.
we have $Y_{1}$={$1$} ,${Y_2}$={$1,-1$},${Y_3}$={$1,\omega,\omega^2$} and so on from this we see that ${Y_2}$ $\nsubseteq$ ${Y_3}$,then no option is true.i am confused here.
Please help
Thankyou
You should be careful when using your intuition of finite unions on infinite unions. Your idea seems to hint that the first one probably isn't a subgroup. Indeed, if we take a $z=\exp(i\frac{2\pi}{99})$ and $w=\exp(i\frac{2\pi}{100})$, then the order of $zw$ is the least common multiple of $99$ and $100,$ which is $9900$, since the two numbers are co-prime. Hence, $zw\not\in \cup_{n=1}^{100} Y_n$, so it's not a sub-group.
For the rest, it's probably worth noting that each of the remaining sets are clearly stable under taking inverses and they all contain $1$. Thus, we simply need to check that for any two elements $z$ and $w$ of the union, $zw$ also belongs to the union.
Using this result, take $z\in Y_{2^n}$ and $w\in Y_{2^m},$ then the order of $zw$ divides $2^{n}\cdot 2^m=2^{n+m}$, so $zw\in Y_{2^{n+m}}$. So $ii)$ is a sub-group.
I think this should give you a fine enough idea of how to handle the last two (it goes fairly similarly).