Answer to Question #145992 in Abstract Algebra for Sourav Mondal

Question

Which of the following statements are true and
which are false ? Give reasons for your answers.
(a) Characteristic of a finite field is zero.
(b) Z12 is a field.
(c) In a ring with unity the sum of any two units
is a unit.
(d) Every element of Sn has order at most n.
(e) There is no non-trivial group homomorphism
from a group of order 5 to a group of order 6.

Answer 1

(a) Characteristic of a finite field is zero.

The characteristic of a finite field $\mathbb Z_2=\{\overline{0},\overline{1}\}$ is 2. Indeed, $2\cdot \overline{0}=\overline{0}+\overline{0}=\overline{0}$, $1\cdot \overline{1}=\overline{1}$ and $2\cdot \overline{1}=\overline{1}+\overline{1}=\overline{2}=\overline{0}$.

Answer: false

(b) $\mathbb Z_{12}$ is a field.

Since $\overline{2}\cdot\overline{6}=\overline{0}, \mathbb Z_{12}$ contains zero divisors. So it is not integral domain, and therefore, it is not a field.

Answer: false

(c) In a ring with unity the sum of any two units is a unit.

Consider the ring $(\mathbb Z,+,\cdot)$. The number $1$ is a unit, but $1+1=2$ is not a unit because there is not integer $x$ such that $2x=1$.

Answer: false

(d) Every element of $S_n$ has order at most $n$.

Consider the symmetric group $S_5$ and its element $a=(123)(45)$. Since the cycles $(123)$ and $(45)$ are independent, the cycles $(123)$ and $(45)$ commute. Taking into account that $|(123)|=3$ and $|(45)|=2$, we conclude that $|a|=6>5.$

Answer: false

(e) There is no non-trivial group homomorphism from a group of order 5 to a group of order 6.

Let $G$ and $H$ be groups, $|G|=5,|H|=5$. Let $a\in G$ be arbitrary. If $f:G\to H$ is a homomorhism, then $|f(a)|$ divides $|a|.$ As a consequence of the Lagrange's theorem, $|a|$ divides $|G|=5$, and therefore, $|f(a)|$ divides 5. On the other hand, $|f(a)|$ divides $|H|=6$. Since the greatest common divisor of 5 and 6 is 1, we conclude that $|f(a)|=1$ for each $a\in G.$ Therefore, $f(a)=e$ for each $a\in G$, where $e$ is identity of $H.$ Consequently, each homomorphism from $G$ to $H$

is trivial.

Answer: true