Group and subgroup of $G$ Let's say a group $G$ and subset $H$ of $G$ are given. Tell whether $H$is a subgroup of $G$ or not.
Given: $G= U(120)$ under modular multiplication, $H = \{a \in G : a^2=1\}$
My work: $H$ is not a subgroup of $G$. 
The identity element of $G$ is $1$.
The inverse of $H$ is not an integer, $(1/a)$
Since the inverse fails, $H$ is not a subgroup
Correction: Could anyone verify if this is true? or show 
 A: I think you are confused on what modular arithmetic is.  It is true that if we were considering the full set of integers $\mathbb{Z}$ under multiplication, this would not be a group due to failing to have inverses for all kinds of numbers.  For example the multiplicative inverse of 2 is $\frac{1}{2}$, which is not an integer.  However $U(n)$ is the group of units, where multiplication is done modulo $n$.  Two numbers are congruent modulo $n$ if their difference is divisible by $n$, or if we can express their difference as a multiple of n. 
Let's take an example with a number smaller than 120, say $U(8)$.  As a set, $U(8)=\{1,3,5,7\}$, the set of numbers between 0 and 7 which are relatively prime to 8.  The reason they need to be relatively prime is to guarantee inverses.  Why don't we care about larger numbers though, like 15?  That's because $15=7 + 8$, and so $15$ is congruent to $7 \mod 8$.  Hence in $U(8)$, $15$ is not any different from $7$.  Similarly, we can take any number larger than 8 and it will be equivalent to a number between $0$ and $7$ by expressing it as multiples of $8$ plus a number between $0$ and $7$.  You can think of this in the same way as the numbers on a clock.  Once you go past $12$, you start back at $1$.  In fact it might be a good idea to add and multiply numbers on a clock so you get familiar with the concept.
Anyway back to $U(8)=\{1,3,5,7\}$.  Let's multiply a few of these numbers together to see what we get.  
Multiplying $3*5=15 = 7 + 1*8$ and so $3*5\equiv 7 (\mod 8)$.
Multiplying $5*7=35=3+4*8$ and so $5*7\equiv 3 (\mod 8)$.
Multiplying $5*5=25=1 + 3*8$ and so $5*5\equiv 1 (\mod 8)$.  
So we see not only is it that $5$ have an inverse in $U(8)$, but it's inverse is itself (as $1 \mod 8$ is the identity of $U(8)$).  So for your question, to say we are defining $H=\{a\in G \vert a^2=e\}$ you are thinking of elements like $5\in U(8)$.  The difference is your group is $U(120)$.  To verify that this set $H$ is a subgroup, you just have to verify that $ab^{-1}\in H$ for $a,b\in H$.  So in this case take two elements $a,b\in H$, compute $(ab^{-1})^2$, and see if you get the identity of $U(120)$.
