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Q)

Which of the following is correct?

\[\](1)An element of a group can have more then one inverse.\[\](2)If every element of a group is its own inverse , then the group is abelian.\[\](3)The set of all $2\times 2$ real matrices forms a group under matrix multiplication.\[\](4)$(a^{\ast} b)^{-1}=a^{-1} \ast b^{-1}$ for all $a,b\in G$

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A)
Statement (1) is false since an element of a group can have only one inverse
Statement (2) is correct.
Statement (3) is false.
Since inverse does not exist.
Singular matrices does not possesinverse
The matrix $\begin {bmatrix} a & a \\a & a \end{bmatrix}$ is a $2 \times 2 $ matrix
Whose inverse does not exist.
Statement (4) is false.
$(a^{\ast} b)^{-1}=b^{-1} \ast a^{-1}$ for all $a,b\in G$
Hence 2 is the correct answer.
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