7 prove that tn f and mathfrakdn f are self normalizing...

Transcript

00:01 Hello everyone so this is the question that we have proved that part a intersection of an arbitrary family of normal subgroups of g is normal in g part b show that if my f tends to g from h is normal is a group of homomorphism and k is normal subgroup then f to the part minus 1 of k is equal to g belonging to all capital g's f of g belongs to all k values basically this means is a normal subgroup normal subgroup of g now part c is we have to deduce this reduce that canal of a group homomorphism is a normal sub and part d is that image of a normal subgroups there show that image of a normal subgroup of a normal subgroup homomorph group homomorph need not be a normal sub so let us jump on to the solution of the question so part a first so we have this claim which is n n of i is normal subgroup of g for e belonging to n n of i for i belonging to i so this implies this is going to be n n of i not equal to phi so let a comma b belonging to n n i for i that it belongs to the integers is going to be equal to a b to the power minus one belonging to n i this implies a b to the part minus one belongs to n n i where i belongs to if we prove it for the subgroup this is going to be let g belongs to g and a belongs to n n i i belongs to i this implies a belongs to n p for all values of i belonging to integer values and multiply both sides with g so this is going to be g a g my to the power minus and belonging to n and i so from this implies we can say that n and i for all values of i belonging to integer is a normal subgroup of g this is the part a of the question now part b of the question so part b f g tending to h so homomorphism m would be f inverse for f to the power minus 1 k m is the normal subgroup of g this is the normal subgroup of g this is the normal subgroup g so we know that f of e would be equal to e dash this is the image of identity so this implies my e belonging to f inverse k would be equal to m that is m would not be equal to now from here i can imply that my m tell g so m is not equal to five so m is delta or the change in the g where it is tending so part c now so part c f g tending to h be a group homomorphism homomorphism group so from the previous value we'll take this this implies this would be k e f not belonging to zero.

05:48 So let us consider a function.

05:51 Consider f -a -b -inverse is equal to f of a multiplied by f of b inverse.

06:00 On solving it what am i gonna get? i'm gonna get e to the power 1 multiplied by e to the power 1.

06:08 So from here i get f of a b inverse is equal to e to the part 1...

Question

7. Prove that $t(n, F)$ and $\mathfrak{d}(n, F)$ are self-normalizing subalgebras of $\mathfrak{gl}(n, F)$, whereas $n(n, F)$ has normalizer $t(n, F)$.

Question

7. Prove that $t(n, F)$ and $\mathfrak{d}(n, F)$ are self-normalizing subalgebras of $\mathfrak{gl}(n, F)$, whereas $n(n, F)$ has normalizer $t(n, F)$.

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