Why use group lasso instead of lasso? I have read the that the group lasso is used for variable selection and sparsity in a group of variables.  I want to know the intuition behind this claim.


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*Why is group lasso preferred to lasso?

*Why is the group lasso solution path not piecewise linear?

 A: Ben's answer is the most general result. But the intuitive answer to the OP is motivated by the case of categorical predictors, which are usually encoded as multiple dummy variables: one for each category. It makes sense in many analyses to consider these dummy variables (representing one categorical predictor) together rather than separately.
If you have a categorical variable with, say, five levels, a straight lasso might leave two in and three out. How do you handle this in a principled manner? Decide to vote? Literally use the dummy variables instead of the more meaningful categorical? How does your dummy encoding affect your choices?
As they say in the introduction of The group lasso for logistic regression, it mentions:

Already for the special case in linear regression when not only continuous but also categorical predictors (factors) are present, the lasso solution is not satisfactory as it only selects individ- ual dummy variables instead of whole factors. Moreover, the lasso solution depends on how the dummy variables are encoded. Choosing different contrasts for a categorical predictor will produce different solutions in general. 

As Ben points out, there are also more subtle links between predictors that might indicate that they should either be in or out together. But categorical variables are the poster child for group lasso.
A: Intuitively speaking, the group lasso can be preferred to the lasso since it provides a means for us to incorporate (a certain type of) additional information into our estimate for the true coefficient $\beta^*$. As an extreme scenario, considering the following:
With $y \sim \mathcal{N} (X \beta^*, \sigma^2 I )$, put $S = \{j : \beta^*_j \neq 0 \}$ as the support of $\beta^*$. Consider the "oracle" estimator $$\hat{\beta} = \arg\min_{\beta} \|y - X \beta\|_2^2 + \lambda \left( |S|^{1/2} \|\beta_S\|_2 + (p-|S|)^{1/2} \|\beta_{S^C}\|_2 \right),$$ which is the group lasso with two groups--one the true support and one the complement. Let $\lambda_{max}$ be the smallest value of $\lambda$ that makes $\hat{\beta} = 0$. Due to the nature of the group lasso penalty, we know that at $\lambda$ moves from $\lambda_{max}$ to $\lambda_{max} - \epsilon$ (for some small $\epsilon > 0$), exactly one group will enter into support of $\hat{\beta}$, which is popularly considered as an estimate for $S$. Due do our grouping, with high probability, the selected group will be $S$, and we'll have done a perfect job. 
In practice, we don't select the groups this well. However, the groups, despite being finer than the extreme scenario above, will still help us: the choice would still be made between a group of true covariates and a group of untrue covariates. We're still borrowing strength.
This is formalized here. They show, under some conditions, that the an upper bound on the prediction error of the group lasso is lower than a lower bound on the prediction error of the plain lasso. That is, they proved that the grouping makes our estimation do better.
For your second question:
The (plain) lasso penalty is piecewise linear, and this gives rise to the piecewise linear solution path. Intuitively, in the group lasso case, the penalty is no longer piecewise linear, so we no longer have this property. A great reference on piecewise linearity of solution paths is here. See their proposition 1. Let $L(\beta) = \|y - X \beta\|_2^2$ and $J(\beta) = \sum_{g \in G} |g|^{1/2} \|\beta_g\|_2$. They show that the solution path of the group lasso is linear if and only if $$\left( \nabla^2L(\hat{\beta}) + \lambda \nabla^2 J(\hat{\beta}) \right)^{-1} \nabla J(\hat{\beta})$$ is piecewise constant. Of course, it isn't since our penalty $J$ has global curvature.
