Math 120A HW 3 Solutions
- Recall that $+_n$ and $\cdot_n$ denote the operations of addition modulo $n$ and multiplication modulo $n$ respectively on the set $\mathbb{Z}_n = \{0,1,\ldots,n-1\}$. They are computed by adding or multiplying normally and then reducing modulo $n$.
-
\begin{array}{c|ccc}
+_4 & 0 & 1 & 2 & 3\\
\hline
0 & 0 & 1 & 2 & 3\\
1 & 1 & 2 & 3 & 0\\
2 & 2 & 3 & 0 & 1\\
3 & 3 & 0 & 1 & 2.
\end{array}
The element $0$ is an identity element for the structure $(\mathbb{Z}_4,+_4)$.
Every element has an inverse in this structure: the inverse of $0$ is $0$, the inverse of $1$ is $3$, the inverse of $2$ is $2$, and the inverse of $3$ is $1$.
-
\begin{array}{c|ccc}
\cdot_4 & 0 & 1 & 2 & 3\\
\hline
0 & 0 & 0 & 0 & 0\\
1 & 0 & 1 & 2 & 3\\
2 & 0 & 2 & 0 & 2\\
3 & 0 & 3 & 2 & 1.
\end{array}
The element $1$ is an identity element for the structure $(\mathbb{Z}_4,\cdot_4)$. The element $0$ has no inverse, the inverse of $1$ is $1$, the element $2$ has no inverse, and the inverse of $3$ is $3$.
- For each of the following objects, say whether or not it is a group. If it is not a group, say why not.
- This binary structure is not a group because there is no identity element. For any element $a$ we have $\max(a,a-1) \ne a-1$, so $a$ is not an identity element.
- Although this binary structure has an identity element, namely $0$,
it is not a group because not every element has an inverse. In fact,
if $a \in [0,1]$ is nonzero then it cannot have an inverse: for all $b \in
[0,1]$ we have $\max(a,b) \ge a \gt 0$, so $\max(a,b)$ is not the identity element, and therefore $b$ is not an inverse for $a$.
- This object is not a group. It is not even a binary structure because $-1,1 \in \mathbb{R}^*$ but $-1 + 1 \notin \mathbb{R}^*$.
- Recall that isomorphisms of binary structures respect identity elements.
In this problem, we show that homomorphisms of groups respect identity elements.
- Let $G$ be a group. If $x = e_G$ then we have $x\ast x = x$ by the defining property of the identity element $e_G$, and conversely, if $x \ast x = x$ then multiplying both sides of this equation on the right by $x^{-1}$ we get the equation $x \ast x \ast x^{-1} = x \ast x^{-1}$, which simplifies to $x = e_G$.
- Let $(G,\ast)$ and $(G',\ast')$ be groups and let $\phi$ be a homomorphism from $(G,\ast)$ to $(G',\ast')$.
We have $e_G \ast e_G = e_G$, and applying $\phi$ to both sides and using the homomorphism property we get $\phi(e_G) \ast' \phi(e_G) = \phi(e_G)$. Therefore $\phi(e_G)$ is the identity element of $G'$ by the previous part.
- Let $(G,\ast)$ be a three-element group, say $G = \{e,a,b\}$ where $e$ is the identity element.
- The symbol that goes in this space is $b$. It cannot be $a$, because there is already an $a$ in that row. If it were $e$, then the symbol to the right of it would have to be $b$, and there would be two $b$'s in the third column, which can't happen either.
-
\begin{array}{c|ccc}
\ast & e & a & b\\
\hline
e & e & a & b\\
a & a & b & e\\
b & b & e & a
\end{array}
-
It is easy to check that the group $(S,\cdot)$ has the property that
$A^2 = I$ for every element $A\in S$ (letting $I$ denote the identity
matrix, which is the identity element for $S$.) This is a structural
property. However, the group $(\mathbb{Z}_4, +_4)$ does not have the
corresponding property: it is not the case that $a +_4 a = 0$ for all $a
\in \mathbb{Z}_4$ (note the change to additive notation.) In particular,
$1 +_4 1 = 2 \ne 0$. Therefore the two groups cannot be isomorphic to
each other.