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Claim: Let $G$ be the set of all real $2 \times 2$ matrices $\left( \begin{array}{cc} a & b \\ 0 & d \end{array} \right)$ such that $ad \not = 0$, with matrix multiplication as the operation. Let $N$ be the subset where $a = d = 1$. Then $N$ is a normal subgroup of $G$.

Showing that $N$ is a subgroup of $G$ is easy because $\left( \begin{array}{cc} 1 & b_1 \\ 0 & 1 \end{array} \right) \left( \begin{array}{cc} 1 & b_2 \\ 0 & 1 \end{array} \right) = \left( \begin{array}{cc} 1 & b_1 + b_2 \\ 0 & 1 \end{array} \right)$. However, I cannot think of a nice way to show that $N$ is a normal subgroup. It would be simple to do out all the computations, but also tedious. Is there a nice way to do this?

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    $\begingroup$ This does not show that $N$ is a subgroup. $\endgroup$ – ThePortakal Sep 23 '15 at 0:11
  • $\begingroup$ It shows $N$ is closed under the operation and has an obvious inverse, which is enough. $\endgroup$ – dalastboss Sep 23 '15 at 0:12
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    $\begingroup$ @dalastboss You also need to show $N$ is non-empty (namely, it has the identity) but that is easy to see as well. $\endgroup$ – Kevin Sheng Sep 23 '15 at 0:24
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You can realize it is the kernel of the map that sends $$\begin{pmatrix}a&b\\0 &c\end{pmatrix}\mapsto \begin{pmatrix} a&0 \\ 0 &c\end{pmatrix}$$

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  • $\begingroup$ Heh, it can hardly get any more elegant than this. +1 $\endgroup$ – guest Sep 23 '15 at 2:54
  • $\begingroup$ Very pretty. Thank you! $\endgroup$ – dalastboss Sep 23 '15 at 3:51
  • $\begingroup$ Upon further investigation, it does not seem to me that this satisfies the homomorphism property $f(xy) = f(x)f(y)$ $\endgroup$ – dalastboss Sep 23 '15 at 5:27
  • $\begingroup$ @dalastboss Did you carry out the multiplication? It does. $\endgroup$ – Pedro Tamaroff Sep 23 '15 at 5:35
  • $\begingroup$ Yeah, you're absolutely right. Chalk that up to arithmetic errors. $\endgroup$ – dalastboss Sep 23 '15 at 5:53
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Note that the eigenvalues of the matrices in $N$ are all $1$. Conjugating by any matrix doesn't change the eigenvalues. Since $G$ is a group, the conjugated matrix must still be upper triangular and have its eigenvalues on the diagonal still, so the diagonal entries must still be 1.

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    $\begingroup$ I lack the linear alg. education to understand this solution, but I will still up vote you $\endgroup$ – dalastboss Sep 23 '15 at 0:14
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This is a clarification of @jgon's answer. To show $N$ is a normal subgroup we want to show that $gNg^{-1} = N$ for any $g \in G$. This is what he means by conjugation (i.e. $gNg^{-1}$). Conjugate matrices have the same eigenvalues since $$\det(gng^{-1}) = \det(g) \cdot \det (n) \cdot \det(g^{-1}) = \det(g)\cdot \det(g^{-1}) \cdot \det(n) = \det (n),$$ and the determinant of a matrix is equal to the product of its eigenvalues. Now since $G$ is a group, $gng^{-1} \in G$ so it must be upper-triangular and have eigenvalues $1$, meaning $gng^{-1} \in N$, precisely what you wanted to prove.

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  • $\begingroup$ Why cannot it have eigenvalues that are inverses of each other, like ${\rm diag}(-1,-1)$? Your argument shows the product of the diagonal elements is $1$. $\endgroup$ – Pedro Tamaroff Sep 23 '15 at 0:29
  • $\begingroup$ Because similar matrices have the same eigenvalues and conjugate matrices are similar by definition. $\endgroup$ – Kevin Sheng Sep 23 '15 at 0:31
  • $\begingroup$ Technically you need that the characteristic polynomials are the same. $\endgroup$ – jgon Sep 23 '15 at 0:34
  • $\begingroup$ @KevinSheng You're talking about eigenvalues when you should be talking about Jordan normal forms. $\endgroup$ – Pedro Tamaroff Sep 23 '15 at 0:57
  • $\begingroup$ Yes but I felt the discussion of eigenvalues is more accessible for the OP. $\endgroup$ – Kevin Sheng Sep 23 '15 at 0:58

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