# Semi-direct product and unitary group: need for subgroups to share only the identity.

The Lie algebra $u(2)$, say, is the direct sum $u(1) \oplus su(2)$, where $u(1)$ has dimension one with generator in the fundamental which we may as well take to be $1\!\!1$. Yet the Lie group $U(2)$ is not isomorphic to the direct product of the Lie groups $U(1) \otimes SU(2)$ where the group elements in $U(1)$ are $e^{i t}1\!\!1$ for $t \in \mathbb{R}$.

Indeed, nor is this group a semi-direct product! The reason usually given is that although the elements of $U(1)$ commute with those of $SU(2)$ (or both are normal subgroups) both subgroups share more than the identity element. Indeed, they both share $-1\!\!1$. To see this, take $t = i\pi$ for $U(1)$ and $\exp(i \pi \hat{n}\cdot \sigma)$ where the Pauli matrices, $\sigma$, are (proportional to) the generators of $su(2)$.

My question is why this is a big deal - certainly in the construction of a direct product $G = G_{1} \otimes G_{2}$ one can decompose any element $(g_{1}, g_{2})$ into $(g_{1}, I_{2})(I_{1}, g_{2})$, a product of elements from the two subgroups. It is also clear that the subgroups $G_{1} = (g_{1}, I_{2})$ and $G_{2} = (I_{1}, g_{2})$ share only the identity, but I struggle to see why this convenience should necessarily be taken forward to the reverse case that one decomposes a group into a direct product.

In particular, for $U(2)$ the $U(1)$ and $SU(2)$ subgroups commute, any element can be decomposed into a product of elements in the two subgroups (even $(-1\!\!1, -1\!\!1) = (1\!\!1, -1\!\!1)(-1\!\!1, 1\!\!1)$) and the product of two elements $(g_{1}, g_{2})(g'_{1}, g'_{2}) = (g_{1}g'_{1}, g_{2}g'_{2})$ is the obvious product that direct product groups are endowed with. The only obstruction seems to be that they share two common elements and I would like to understand why this is a problem.

• What's the actual question? May 13, 2017 at 18:43
• Why would the two commuting (or normal) subgroups sharing more than just the identity element be a problem. Or reversed, why does the definition of a direct or semi-direct product require that the two subgroups share only the identity.
– lux
May 13, 2017 at 18:44
• I'm sorry I still don't see what the problem is. May 13, 2017 at 18:46
• Another way of asking the same question: Why could I not define the direct product (or the semi-direct product) without the requirement that the two subgroups only share the identity. What would go wrong if I were to do that?
– lux
May 13, 2017 at 18:48
• Why should you want to redefine the term direct product in such a peculiar and confusing way? May 13, 2017 at 18:50

I think I've answered my own question - thank you to the community for offering useful suggestions.

If one wishes to try to decompose $G$ into two subgroups $G_{1}$ and $G_{2}$ and state that $G$ is isomorphic to the product group $G_{1} \otimes G_{2}$ with the usual product $(g_{1}, g_{2})(g'_{1}, g'_{2}) = (g_{1}g'_{1}, g_{2}g'_{2})$ then it must be the case that $G_{1} \cap G_{2} = 1\!\!1$ simply because every element of $G_{1} \otimes G_{2}$ can be written as $(g_{1}, g_{2}) = (g_{1}, 1\!\!1)(1\!\!1, g_{2})$.

Since $U(2)$ does not satisfy this requirement it cannot possibly be isomorphic to $U(1) \otimes SU(2)$ nor $U(1) \ltimes SU(2)$.

Let $U$ and $V$ be groups, and $Z(\_)$ denotes the center. The following is exercise $7.9$ in Isaac's algebra.

Let $M \subset Z(U)$ and $N \subset Z(V)$ and assume $M \cong N$. Then there is a group $G$ with normal subgroups $K$ and $L$ so that $K \cong U$ and $L \cong V$ and $KL = G$ and $[K,L] = 1$ (they commute) and $K \cap L = M$ (after identifying $K$ with $U$).

This is called the central product of $U$ and $V$ identifying $M$ and $N$.

Note that if you have two subgroup $G_1$ and $G_2$ of $G$, and you insist that they commute, then $G_1 \cap G_2$ is in the center of both of them.