Exercise · ZW-FIRST-ISOMORPHISM-THEOREM-002 · ★☆☆ Basic

The determinant \(\det:GL_2(\mathbb{R})\to\mathbb{R}^*\) is a surjective group homomorphism with kernel \(SL_2(\mathbb{R})\). What does the first isomorphism theorem state?

Choose one option.

Show answer
  • A. \(GL_2(\mathbb{R})/SL_2(\mathbb{R})\cong\mathbb{R}^*\)

Explanation

Since \(\det\) is surjective, its image is all of \(\mathbb{R}^*\), and the theorem gives the isomorphism between the quotient by the kernel and the image. It does not say the original group is isomorphic to the kernel.

Study Modern algebra I in ZWMath →

Review the concept: First isomorphism theorem

Every group homomorphism \(\varphi:G\to H\) induces an isomorphism between the quotient of \(G\) by the kernel of \(\varphi\) and the image of \(\varphi\).

\[G/\ker\varphi\;\cong\;\varphi(G),\qquad a\ker\varphi\mapsto\varphi(a)\]

More practice problems: first isomorphism theorem

← ZW-FIRST-ISOMORPHISM-THEOREM-001 ZW-FIRST-ISOMORPHISM-THEOREM-003 →