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Please determine order The automorphisms group of the group $Z_{8}\times Z_{4}$, where $Z_{8}$ and $Z_{4}$ are cyclic groups of order $8$ and $4$, respectivly.

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As a new user to Math.SE I would like to tell you that when posting questions here, it is considered impolite to instruct other people to solve your questions for you. Please provide some context, what you have tried, where you are stuck etc. – user38268 May 19 '12 at 6:38
I think the order of it is $2^7$. Do it is true? – ali May 19 '12 at 6:41
Plz, show your attempt for solving the problem. What have you done for it? – Babak S. May 19 '12 at 7:21
@ali: The order must be even, since the group is an abelian one. – Babak S. May 19 '12 at 8:13
@ali: See this paper AUTOMORPHISMS OF FINITE ABELIAN GROUPS by CHRISTOPHER J. HILLAR AND DARREN L. RHEA. The last Theorem give you the order of the group. The order is $2^7$. – Babak S. May 19 '12 at 9:14

An endomorphism of $\mathbf{Z}_8\times\mathbf{Z}_4$ is completely determined by the images of $(1,0)$ and of $(0,1)$. We can view such a map as a $2\times 2$ matrix, identifying the map $$\begin{align*} (1,0) &\mapsto (a,b)\\ (0,1) &\mapsto (c,d) \end{align*}$$ with the matrix $$\left(\begin{array}{cc} a & c\\ b & d \end{array}\right)$$ where the first row is taken modulo $8$, and the second row is taken modulo $4$.

Note that since the image of $(0,1)$ must have exponent $4$, it follows that $c$ must in fact be a multiple of $2$, so we can in fact rewrite $(c,d)$ as $(2c,d)$, and then take $a$ modulo $8$, and $b,c,d$ all modulo $4$.

Composition of maps corresponds to multiplication of matrices. Note that the multiplication is well defined under the condition that the (1,2) entry be a multiple of $2$, since $$\left(\begin{array}{cc} x & 2z\\ y & w \end{array}\right) \left(\begin{array}{cc} a & 2c\\ b & d \end{array}\right) = \left(\begin{array}{cc} ax +2bz & 2(cx+dz)\\ ay+bw & 2cy+wd \end{array}\right).$$ Although $b$ is only defined modulo $4$, the quantity $2bz$ is well defined modulo $8$, since if $b'=b+4k$, then $2b'z = 2bz + 8kz\equiv 2bz\pmod{8}$.

So the total number of endomorphisms is equal to the total number of matrices with these properties. There are $8$ choices for $a$, and $4$ choices each for $b$, $c$, and $d$. This gives $2^9$ endomorphisms.

The automorphisms are the ones that are invertible. In order for the matrix to be invertible, we need the determinant to be invertible, which means the determinant must be odd. The determinant of $$\left(\begin{array}{cc} a & 2c\\ b & d \end{array}\right)$$ is $ad-2bd$, so we need $ad$ to be odd; that is, $a$ and $d$ must both be odd. You should now verify that if $a$ and $d$ are both odd, then we indeed have an invertible homomorphism.

Alternatively, one can also view an endomorphism of $\mathbf{Z}_8\times\mathbf{Z}_4$ as a quadruple of maps, $(\phi,\psi,\rho,\sigma)$, where $\phi\colon\mathbf{Z}_8\to\mathbf{Z}_8$, $\psi\colon\mathbf{Z}_8\to\mathbf{Z}_4$, $\rho\colon\mathbf{Z}_4\to\mathbf{Z}_8$, and $\tau\colon\mathbf{Z}_4\to\mathbf{Z}_4$, using the universal properties of the finite product of abelian group (both a product and a coproduct in the category). In order to be an automorphism, we need $\phi$ and $\rho$ to be invertible, and one can verify that this is sufficient as well. This gives you an easy way to count them, giving the same values as above.

How many are there? There are now $4$ choices for $a$, $4$ choices for $c$, $4$ choices for $b$, and $2$ choices for $d$, which gives $2^7$ automorphisms.

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Thank you prof Magidin – ali May 22 '12 at 14:07

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