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I see here that one can prove that $$ SL_2(\mathbb{Z}_5) / \{\pm I\} \simeq A_5 $$ using the First Isomorphism Theorem.

My question is how one would do that.

I know that I need a surjective homomorphism $$ T: SL_2(\mathbb{Z}_5) \to A_5 $$ with kernel $\{\pm I\}$. The only homomorphism I have come across with matrix groups is the determinant map, so my question is what homomorphism would work here.

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  • $\begingroup$ To use the isomorphism theorem, which states $G/\operatorname{ker}\phi\cong\operatorname{im}\phi$, you don't need a surjective homomorphism to $A_5$ but a surjective homomorphism to any group with kernel isomorphic to $A_5$. $\endgroup$ – theage Apr 30 '15 at 19:50
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    $\begingroup$ The case $p= 5$ is a bit trickier than, say, $p = 7$ (and I'm not aware of any natural way in which the group acts on a set of 5 points). Try considering $PSL_2(\mathbb{Z}_5) = SL_2(\mathbb{Z}_5)/\{\pm 1\}$ as the group of symmetries of a regular icosahedron. Alternatively, $PSL_2(\mathbb{Z}_5)$ and $A_5$ are both simple groups of order $60$, and there's exactly one such group. (There are some well-known but nontrivial facts in that previous sentence.) $\endgroup$ – anomaly Apr 30 '15 at 20:03
  • $\begingroup$ @anomaly: How does it work for $p=7$? $\endgroup$ – John Doe Apr 30 '15 at 20:04
  • $\begingroup$ The group $PSL_2(\mathbb{Z}_7) = GL_3(\mathbb{Z}_2)$ conveniently acts on the space $\mathbb{P}^2(\mathbb{Z}_2)$, which has $7$ points. (It's not isomorphic to $A_7$, though.) $\endgroup$ – anomaly Apr 30 '15 at 20:34
  • $\begingroup$ cf. math.stackexchange.com/q/93762/152 $\endgroup$ – Grigory M May 13 '15 at 12:57
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The group $PGL(2,\mathbb{F}_5)$ acts faithfully on the set $X$ of lines of the vector space $\mathbb{F}_5^2$. One can show that $X$ has cardinal $6$. Hence the action gives an injective group morphism $\rho$ from $PGL(2,\mathbb{F}_5)$ to $\mathfrak{S}_6$ the symmetric group on $6$ elements.

Now $\rho(PGL(2,\mathbb{F}_5))$ is of cardinal$120$ hence of index $6$ in $\mathfrak{S}_6$ and hence (group theory behind this) isomorphic to $\mathfrak{S}_5$.

Finally $PSL(2,\mathbb{F}_5)$ is of index $2$ in $PGL(2,\mathbb{F}_5)$ and hence isomorphic to $\mathfrak{A}_5$.

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  • $\begingroup$ So this isn't really a good example of the application of the First Isomorphism Theorem? $\endgroup$ – John Doe May 1 '15 at 11:48
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    $\begingroup$ If I've understood well, the OP is asking for a proof using the fundamental isomorphism theorem for groups. $\endgroup$ – user26857 May 16 '15 at 9:01
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Let $G=SL_2(\mathbb{Z}_5)$, $S$ be a Sylow $2$-subgroup of $G$. Then $[G:N_G(S)]=5$, hence we have $5$ Sylow $2$-subgroups of $G$, and $G$ acts transitively on them. In such a way we get transitive permutation representation $\rho:G\to S_5$ with $\ker(\rho)=\{\pm I\}$. Since $G$ has not a subgroups of index $2$, containing $\{\pm I\}$, then $\rho(G)\leq A_5$. But $|\rho(G)|=|A_5|$, hence we have epimorphism $\rho:G\to A_5$ and by the fundamental theorem on homomorphisms $SL_2(\mathbb{Z}_5)/\{\pm I\}\cong A_5$.

How to show, that $[G:N_G(S)]=5$? Note, that $|G|=2^3\cdot 3\cdot 5$, hence each Sylow $2$-subgroup of $G$ has order $8$. For natural ring's homomorphism $\mathbb Z\to\mathbb{Z}_5$ we will use a bar convention. Let $$ A= \begin{pmatrix} \bar 2 & \bar 0 \\ \bar 0 & -\bar 2 \end{pmatrix}, B=\begin{pmatrix} \bar 0 & \bar 1 \\ -\bar 1 & \bar 0 \end{pmatrix}. $$ Then $A,B\in G$ and $$ A^2=B^2=-I, B^{-1}AB=A^{-1}, o(A)=o(B)=4. $$ Hence $S:=\langle A,B\rangle\simeq\mathbb{Q}_8$ and $|S|=8$. Therefore $S$ is a Sylow $2$-subgroup of $G$. It is known that quaternion group has $3$ subgroups of order $4$. In $S$ subgroups of order $4$ are $\Omega:=\{\langle A\rangle,\langle B\rangle,\langle C\rangle\}$, where $C=AB$. If $X\in N:=N_G(S)$, then $\langle X\rangle$ acts on $\Omega$ by conjugation. This suggests try to find such $X\in G$, that $\langle X\rangle$ acts on $\Omega$ transitively. We may try to find, for example, such $X$, that $A^X=C$ and $B^X=-A$ (it comes down to solving a very simple systems of equations). I did these easy calculations and found one such matrix: $$ X= \begin{pmatrix} \bar 2 & \bar 2 \\ \bar 1 & -\bar 1 \end{pmatrix}. $$ It follows that $3\shortmid|N|$. If $5\shortmid|N|$, then $S\unlhd G$, it is impossible, since $PSL_2(5)$ simple (in fact, we can avoid the use of simplicity here). In such a way $|N|=2^3\cdot 3$ and $[G:N_G(S)]=5$.

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  • $\begingroup$ How do you know that there are 5 Sylow-2-subgroup of G and not 3 or 15? $\endgroup$ – Stanley May 20 '15 at 1:22
  • $\begingroup$ @SimonRigby I expanded my answer. $\endgroup$ – Alex W May 20 '15 at 9:39
  • $\begingroup$ @AlexW Well, proving that number of 2-Sylow subgroups of SL_{2}(\mathbb{Z}_{5}}$ is 5 looks extremely hard. Is there an easier method? $\endgroup$ – crskhr Apr 25 '17 at 17:30
  • $\begingroup$ @S.C. In principle all what we need is the existence of subgroup $M\leq G$ of index 5, and we can define $M:=\langle S,X\rangle$ and proof that $|M|=2^3\cdot 3$ - it is easy, then $[G:M]=5$ and $G$ acts transitively on the 5 cosets of $G$ by $N$. Hence we get transitive permutation representation ... . In the previous arguments we get more information than we need - equality $N_G(S)=M$. Sylow theorem is also unnecessary. $\endgroup$ – Alex W Apr 26 '17 at 15:40

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