# A problem on normal subgroups of the general linear group

Let $GL_n(\mathbb{R})$ denote the group of all $n×n$ matrices with real entries (with respect to matrix multiplication) which are invertible. Pick out the normal subgroups from the following:
a. The subgroup of all real orthogonal matrices.
b. The subgroup of all invertible diagonal matrices.
c. The subgroup of all matrices with determinant equal to unity.

I get that (c) is true. but not sure about others but guess that they are not true.I have try to get some counterexamples but failed.so please provide me some counterexamples.

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 The subgroup of $c)$ is clearly normal because its the kernel of the homomorphism that sends each matrix to its determinant. – Amr Jan 7 at 4:20

(c) is clearly normal subgroup as for any $g\in GL_n(\mathbb{R})$ $\det(ghg^{-1})=\det(g)\det(h)\det(g^{-1})=\det(g)\cdot 1\cdot(\det g)^{-1}=1$.

(a) to be normal subgroup we need $ghg^{-1}(ghg^{-1})^t=I$ but that need not be true for any $g\in GL_n({\mathbb R})$ just check yourself.

(b)clearly not normal, take any matrix from $GL_n(\mathbb R)$ whose atleast one entry is $0$ in first collumn, what do get when multply it with the diagonal matrix?

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A diagonal matrix is invertible iff there's no zero entry on its main diagonal, so (b) is clearly normal, just as (c) is, and (a) isn't as remarked by Kuttus.

Added: The above is wrong, of course: take any non-diagonal diagonalizable matrix $\,A\,$ , so that there exists an invertible matrix $\,P\,$ and a diagonal one $\,D\,$ s.t.:

$$P^{-1}AP=D\Longrightarrow A=PDP^{-1}\Longrightarrow$$

the subgroup of diagonal matrices is not normal!

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 I really do not getting sure the fact that $ghg^{-1}$ will be again a invertible diagonal matrix for every $g$, could you give me some hint? – Taxi Driver Jan 7 at 5:39 the given answer is only (c). that's why I am little confused. – ranadip ganguly Jan 7 at 7:34 Well, I may have made a mistake, and in fact I did, but it is odd to see how people rush to downvote instead of trying to either to correct or to ask for clarifications. BTW, affirming without proof that (c) is clearly not normal seems to be in the same level as my claim, so we haven't yet advanced, but right now I'm going to add an explanation to my answer, correcting it. – DonAntonio Jan 7 at 11:33