A. Recognizing Subgroups
In parts 1-6 below, determine whether or not $H$ is a subgroup of
$G$. (Assume that the operation of $H$ is the same as that of $G$.)
Instructions If $H$ is a subgroup of $G$, show that both
conditions in the definition of "subgroup" are satisfied. If $H$ is not a
subgroup of $G$, explain which condition fails.
Example $G = *$, the multiplicative group of the real
numbers.
$H = \{2^n : n \in \mathbb{Z} \}$. $H$ is not a subgroup of $G$.
(i) If $2^n$, $2^m \in H$, then $2^n 2^m = 2^{n+m}$. But $n + m \in \mathbb{Z}$, so $2^{n+m} \in H$.
(ii) If $2^n \in H$, then $1/2^n = 2^{-n}$. But $-n \in \mathbb{Z}$, so $2^{-n} \in H$.
(Note that in this example the operation of $G$ and $H$ is
multiplication. In the next problem, it is addition.)
1 $G = \mathbb{R}$, $+$, $H = \{\log a : a \in \mathbb{R}, a > 0\}$. $H$ is not a
subgroup of $G$.
2 $G = \mathbb{R}$, $+$, $H = \{\log n : n \in \mathbb{N}, n > 0\}$. $H$ is not a
subgroup of $G$.
3 $G = \mathbb{R}$, $+$, $H = \{x \in \mathbb{R} : \tan x \in \mathbb{Z}\}$. $H$ is not a
subgroup of $G$.
HINT: Use the following formula from trigonometry:
4 $G = \mathbb{R}$, $*$, $H = \{2^{n3^m} : m, n \in \mathbb{Z}\}$. $H$ is not a
subgroup of $G$.
