# Help in this little doubt in this proof from Hoffman and Kunze's Linear Algebra book

I'm reading Hoffman and Kunze's Linear Algebra book and on page 177 they stated and proved the following theorem:

It's a big proof which I didn't understand only a very little part of it:

I tried a lot but I couldn't understand why $\sigma_2^{-1}\gamma\sigma_1^{-1}\in G(r+s,t)$.

I think this fact is not from advanced group theory and we need only a basic knowledge of abstract algebra to prove this. Therefore I'm almost sure I missed something silly or the authors made a mistake in this proof.

I need help.

Thanks

• The $\sigma$s by definition leave the members of the index set $I_t=\{r+s+1,\ldots,r+s+t\}$ invariant and $\gamma$ by definition permute the members of $I_t$ among themselves. Therefore $\sigma_2\gamma\sigma_1^{-1}$ permute the members of $I_t$ among themselves.
• Similarly, by definition, the $\sigma$s and $\gamma$ permute the members of the index set $I_{rs}=\{1,2,\ldots,r+s\}$ among themselves. Therefore $\sigma_2\gamma\sigma_1^{-1}$ also permute the members of $I_{rs}$ among themselves.
• So, in summary, $\sigma_2\gamma\sigma_1^{-1}$ permute the members of $I_{rs}$ among themselves and it also permute the members of $I_t$ among themselves. Now that's exactly what a permutation in $G(r+s,t)$ means.
• Why is the $\gamma$ an identity on $I_{rs}$? thank you very much for your answer! Commented Jul 4, 2016 at 7:47
• @user42912 Sorry, that was wrong. $\gamma$ by definition is in $G(r,s,t)$. It isn't an identity on $I_{rs}$. Nonetheless it permutes the members of $I_{rs}$ among themselves. This is now fixed in the answer. Commented Jul 4, 2016 at 7:52