Math 120A HW 3 Solutions

  1. 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 $n$.
    1. \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 $1$.
    2. \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 $3$.
  2. For each of the following objects, say whether or not it is a group. If it is not a group, say why not.
    1. 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.
    2. 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 $a$.
    3. This object is not a group. It is not even a binary structure because $-1,1 \in \mathbb{R}^*$ but $-1 + 1 \notin \mathbb{R}^*$.
  3. Recall that isomorphisms of binary structures respect identity elements. In this problem, we show that homomorphisms of groups respect identity elements.
    1. 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 $x = e_G$.
    2. 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.
  4. Let $(G,\ast)$ be a three-element group, say $G = \{e,a,b\}$ where $e$ is the identity element.
    1. 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.
    2. \begin{array}{c|ccc} \ast & e & a & b\\ \hline e & e & a & b\\ a & a & b & e\\ b & b & e & a \end{array}
  5. 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.