Let G be a group of $2\times 2$ non-singular matrices under multiplication over the field $\mathbb{Z}_3$

Question

Let $G$ be a group of $2\times 2$ non-singular matrices under multiplication over the field $\mathbb{Z}_3$. Define a map $f:G \to \mathbb{R}^*$ (where $\mathbb{R}^*$ is a group of non zero real under multiplication) s.t. $f(A)= |A|$. Then $o(\ker f)$ is

a) $12$

b) $24$

c) $48$

d) none of the above

O(ker f) will be those element of gG which is mapped to identity matrices over Z3 — Halima.Khatun, Dec 30 '16 at 04:49
@Halima.Khatun; $\ker f={A:|A|=1}$ where $|A|=$ order of $A$.So then $\ker f={I}$ where $I$ is identity — Learnmore, Dec 30 '16 at 05:02
@BenStokes I think $|A|$ is intended to mean determinant, not order. — Robert Israel, Dec 30 '16 at 05:05

score 0 · Answer 1 · edited Apr 13 '17 at 12:21

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Hint: Ker$(f)$ is nothing but $SL_2(\mathbb Z_3)$,Can you compute the cardinality of $SL_2(\mathbb Z_3)$?

Edit: I think you know how to compute the order of $GL_n(\mathbb F_p)$,to compute the order of $SL_n((\mathbb F_p)$ note that there is a 'natural' onto group homomorphism $f: GL_n((\mathbb F_p) \to \mathbb F_p^*$ (What is the map?) and hence apply the first isomorphism theorem to compute the cardinality of $SL_n((\mathbb F_p)$... Still if you don't get it you can see here.

edited Apr 13 '17 at 12:21

Community

1

answered Dec 30 '16 at 04:52

Arpit Kansal

10,268

no...plz help me – Halima.Khatun Dec 30 '16 at 05:01
$\ker f={A:|A|=1}$ where $|A|=$ order of $A$.So then $\ker f={I}$ where $I$ is identity – Learnmore Dec 30 '16 at 05:04
Am I missing something?Where from do you get $SL_2$ – Learnmore Dec 30 '16 at 05:05
Determinant, not order. Order is not a homomorphism. – Robert Israel Dec 30 '16 at 05:06
Dear @BenStokes: Recall that $o(ab) \neq o(a)o(b)$ unless there are assumptions on $G$ like abelian etc..Hence order can't be homomorphism as pointed out by Robert Israel. – Arpit Kansal Dec 30 '16 at 05:10

score 0 · Answer 2 · answered Dec 30 '16 at 04:59

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A member of $\ker f$ is a matrix $\pmatrix{a & b\cr c & d\cr}$ with $a,b,c,d \in \mathbb Z_3$ and $ad - bc \equiv 1 \mod 3$.
For any $a \ne 0$, and any $b$ and $c$, you need $d \equiv (1+bc)/a \mod 3$. That makes $2 \times 3 \times 3 = 18$ choices. On the other hand, if $a = 0$ you can take any $d$, but you need $bc=1$, thus either $b=c=1$ or $b=c=2$. This makes $6$ additional choices, for a total of $24$.

answered Dec 30 '16 at 04:59

Robert Israel

448,999

thanks..do there have any general rule to compute such kind of problem – Halima.Khatun Dec 30 '16 at 05:07

Let G be a group of $2\times 2$ non-singular matrices under multiplication over the field $\mathbb{Z}_3$

2 Answers2