Question

Which of the following is a group?

• A. $\{1,2,4,8\}$ under multiplication
• B. $\{0,\pm 2,\pm 4,\pm 6...\}$ under addition
• C. $\{1,-1\}$ under addition
• D. $\{0,1,2,3,4\}$ under multiplication modulo $5$

Answer 1

Answer: (B)

$\{0,\pm 2,\pm 4,\pm 6...\}$ under addition

A)

$G\equiv \{1,2,4,8\}$ under multiplication

If we multiply elements of the above set, then it is not necessary that it will lie in the same set.

$1\times 2=2\in G$

$2\times 4=8\in G$

$4\times 8=32\notin G$

Not a group.

B)

$H\equiv \{0,\pm 2,\pm 4,\pm 6,\cdots \}$ under addition

If we add any elements of this set, then we will get the result in the same set.

$0+2=2\in H$

$0-2+4-6=-4\in H$ and so on,

It is a group.

C)

$F\equiv \{1,-1\}$ under addition

If we add any elements of this set, then it is not necessary that it will lie in the same set.

$1+-1=0\notin F$

Not a group.

D)

$P\equiv \{0,1,2,3,4\}$ under multiplication modulo $5$

Step 1: Check that the operation defined in the question is a binary operation:

Making Cayley table, we get:

	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$
$2$	$0$	$2$	$4$	$1$	$3$
$3$	$0$	$3$	$1$	$4$	$2$
$4$	$0$	$4$	$3$	$2$	$1$

Calculations above show that the operation defined in the question assigns to each ordered pair of elements of $\{0,1,2,3,4\}$ an element in $\{0,1,2,3,4\}$. Hence, it is a binary operation.

Step 2: Check the existence of identity:

If we look at the Cayley table above, we see that $1.a=a.1=a$ for all $a$ in $\{0,1,2,3,4\}$. So, $1$ is an identity.

Step 3: Check the existence of inverses:

The Clearly table also shows that for each element $a$in $\{0,1,2,3,4\}$, there is an element $b$in $\{0,1,2,3,4\}$ such that $a\cdot b=b\cdot a=e$.

Except element $0$

$1\cdot 1=1\cdot 1=1$

$2\cdot 3=3\cdot 2=1$

$3\cdot 2=2\cdot 3=1$

$4\cdot 4=4\cdot 4=1$

But

$0\cdot b=b\cdot 0\ne 1$

Hence, not satisfy inverse property.

Therefore, it is not a group.