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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $G = \frac{R}{{0}}$ and $H = {-1, 1}$ be groups under the multiplication. Then, the map $\phi : G \to H$ defined by $\phi(x) = \frac{x}{|x|}$ is
Solution
$\phi(x) = \frac{x}{|x|}$ gives the sign of $x$.
If $x > 0$, then $\phi(x) = 1$
If $x < 0$, then $\phi(x) = -1$
Also,
$\phi(xy) = \phi(x)\phi(y)$
So, it is a homomorphism.
It is onto because both $1$ and $-1$ are obtained.
It is not one-one because many positive numbers map to $1$, and many negative numbers map to $-1$.
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