WEll, no, because neither group is abelian. While the translation group IS abelian, the orthogonal group is NOT:
consider $A,B \in O_2$:
$\displaystyle A = \begin{bmatrix}\frac{1}{2}&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&\frac{1}{2} \end{bmatrix}$
$B = \begin{bmatrix}1&0\\0&-1 \end{bmatrix}$
Then:
$\displaystyle AB = \begin{bmatrix}\frac{1}{2}&\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&-\frac{1}{2} \end{bmatrix}$
while:
$\displaystyle BA = \begin{bmatrix}\frac{1}{2}&-\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}&-\frac{1}{2} \end{bmatrix}$
In general, if $G$ is a direct product of two subgroups $H,K$, one most show 3 things:
1) $H,K$ are both NORMAL in $G$
2) $G = HK$
3) $H \cap K = \{1_G\}$
Presumably, you have already shown (2) and (3) hold. The problem, therefore, must lie with (1).
Yes, IF their centers are not isomorphic, the groups are not isomorphic. However, the converse is NOT true: you cannot conclude that two groups with isomorphic centers are isomorphic. Determining $Z(M_n)$ might not be as easy as you think, however.
$M_n$ is "almost" a direct product of $O_n$ and $T_n$. We can let the orthogonal group act on the translations like so:
"rotate/flip", translate, then "rotate back/unflip", which amounts to a DIFFERENT translation (translation by the reverse rotated vector, or the flipped vector, instead of our original one). And we have:
$O_n \cong M_n/T_n$
which means that isometries are orthogonal maps "up to a translational factor". The problem is that this action of the orthogonal group isn't *trivial*, we do not, in general, preserve our original translation (just its orientation).
For example, consider the translation by (3,1) in $\Bbb R^2$. Let's say we rotate by 90 degrees counter clockwise first.
This sends (x,y) to (-y,x). Now we translate, and we have (3-y,x+1). The reverse rotation sends (a,b) to (b,-a). Applying this to (3-y,x+1), we wind up with the point (x+1,y-3), so the overall effect is to translate by (1,-3) (which is where the "reverse rotation" sends (3,1)).
The problem here is, when you translate, perform some orthogonal map, and then "translate back", you typically do NOT preserve the origin. Let's use our previous example:
First we translate, sending (x,y) to (x+3,y+1). Rotating the plane by 90 degrees counterclockwise now sends this point to (-y-1,x+3). Translating back (by (-3,-1)), we wind up with (-y-4,x+2), or:
$t_{(-4,2)} \circ p_{\pi/2}$
which sends the origin to (-4,2) (and thus cannot be a linear map).
It turns out that we have a semi-direct product, not a direct product. When we compose the product of a translation and an orthogonal map (an isometry) with another such isometry, the orthogonal part of the second transformation "garbles (re-arranges)" the translational part of the first isometry, they don't act independently.
Again, a more in-depth look:
For our first isometry, we'll use $M_1 = t_{(3,1)} \circ p_{\pi/2}$, and for our second one, we'll use $M_2 = t_{(2,2)} \circ r$ where $r$ is reflection about the x-axis.
IF we had a direct product, we would expect to get $t_{(5,3)} \circ r'$, where $r'$ has the matrix:
$\begin{bmatrix}0&-1\\-1&0 \end{bmatrix}$.
That is, we would expect $M_2 \circ M_1 (x,y) = (5-y,3-x)$.
Let's see what we actually get:
$M_1(x,y) = t_{(3,1)}(p_{\pi/2}(x,y)) = (3-y,x+1)$
$M_2(x,y) = t_{(2,2)}(r(x,y)) = (x+2,2-y)$, so:
$M_2 \circ M_1 (x,y) = M_2(3-y,x+1) = (5-y,1-x) = t_{(5,1)} \circ r'(x,y)$
Why do we wind up translating by (5,1) instead of (5,3)?
Let's look at what happens to the origin:
The first isometry moves the origin to (3,1). But when we reflect (3,1) about the x-axis, we get (3,-1), which winds up at (5,1) when we translate by (2,2). The reflection "messes up" our translational part. Note that the "orthogonal part" is unaffected by the translation, the matrix we get in both cases is still $r'$ (we still wind up swapping x with y and changing the sign).
This kind of "tangled interaction" is typical of semi-direct products.