Question #145992
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.
1
Expert's answer
2020-11-24T17:29:22-0500

(a) Characteristic of a finite field is zero.


The characteristic of a finite field Z2={0,1}\mathbb Z_2=\{\overline{0},\overline{1}\} is 2. Indeed, 20=0+0=02\cdot \overline{0}=\overline{0}+\overline{0}=\overline{0}, 11=11\cdot \overline{1}=\overline{1} and 21=1+1=2=02\cdot \overline{1}=\overline{1}+\overline{1}=\overline{2}=\overline{0}.


Answer: false


(b) Z12\mathbb Z_{12} is a field.


Since 26=0,Z12\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 (Z,+,)(\mathbb Z,+,\cdot). The number 11 is a unit, but 1+1=21+1=2 is not a unit because there is not integer xx such that 2x=12x=1.


Answer: false


(d) Every element of SnS_n has order at most nn.


Consider the symmetric group S5S_5 and its element a=(123)(45)a=(123)(45). Since the cycles (123)(123) and (45)(45) are independent, the cycles (123)(123) and (45)(45) commute. Taking into account that (123)=3|(123)|=3 and (45)=2|(45)|=2, we conclude that a=6>5.|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 GG and HH be groups, G=5,H=5|G|=5,|H|=5. Let aGa\in G be arbitrary. If f:GHf:G\to H is a homomorhism, then f(a)|f(a)| divides a.|a|. As a consequence of the Lagrange's theorem, a|a| divides G=5|G|=5, and therefore, f(a)|f(a)| divides 5. On the other hand, f(a)|f(a)| divides H=6|H|=6. Since the greatest common divisor of 5 and 6 is 1, we conclude that f(a)=1|f(a)|=1 for each aG.a\in G. Therefore, f(a)=ef(a)=e for each aGa\in G, where ee is identity of H.H. Consequently, each homomorphism from GG to HH

is trivial.


Answer: true



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