Use group by inside subquery, MySQL

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#MySQL 5.7.12
#please drop objects you've created at the end of the script 
#or check for their existance before creating
#'\\' is a delimiter

select version() as 'mysql version';
drop table house_details;
CREATE TABLE IF NOT EXISTS `house_details` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `date` varchar(255) DEFAULT NULL,
  `receipt_id` varchar(50) NOT NULL,
  `ward` varchar(200) DEFAULT NULL,
  `house_id` varchar(255) DEFAULT NULL,
  `name` varchar(255) DEFAULT NULL,
  `rupees` int(50) DEFAULT NULL,
  `paid` varchar(255) DEFAULT NULL,
  `outstanding` varchar(255) DEFAULT NULL,
  `financial_year` varchar(100) DEFAULT NULL,
  `income` varchar(100) DEFAULT NULL,
  `subincome` varchar(200) DEFAULT NULL,
  `ssubincome` varchar(200) DEFAULT NULL,
  `createdAt` datetime NOT NULL,
  `updatedAt` datetime NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=307 DEFAULT CHARSET=latin1;

--
-- Dumping data for table `house_details`
--

INSERT INTO `house_details` (`id`, `date`, `receipt_id`, `ward`, `house_id`, `name`, `rupees`, `paid`, `outstanding`, `financial_year`, `income`, `subincome`, `ssubincome`, `createdAt`, `updatedAt`) VALUES
(288, '2017-02-22', '288', 'ward 1', '22', 'Kannan', 300, NULL, NULL, NULL, 'Taxes', 'Garbage tax', NULL, '2017-11-22 10:37:28', '2017-11-23 10:37:28'),
(301, '2017-09-05', '301', '1', '2', 'Vignesh', 4000, NULL, NULL, NULL, 'Grants from government', 'Special Grants', 'Matching grant', '2017-12-05 10:57:11', '2017-12-05 10:57:11'),
(302, '2017-12-05', '302', '1', '2', 'Vignesh', 236, NULL, NULL, NULL, 'Grants from government', 'General Grants', 'Grants in Lieu of Octroi', '2017-12-05 10:59:40', '2017-12-05 10:59:40'),
(303, '2017-12-06', '303', '2', '22', 'Bala', 233, NULL, NULL, '2017-2018', 'Taxes', 'Light tax', NULL, '2017-12-06 12:30:14', '2017-12-06 12:30:14'),
(304, '2017-12-06', '304', '2', '22', 'Bala', 445, NULL, NULL, '2013-2014', 'Taxes', 'Hoarding tax', NULL, '2017-12-06 12:30:39', '2017-12-06 12:30:39'),
(305, '2017-12-07', '305', '3', '44', 'raja', 104, NULL, NULL, NULL, 'Taxes', 'Light tax', 'Matching grant', '2017-12-07 05:01:05', '2017-12-07 05:01:05'),
(306, '2017-12-07', '306', '4', '45', 'Rahul', 500, NULL, NULL, NULL, 'Taxes', 'Light tax', NULL, '2017-12-07 05:16:43', '2017-12-07 05:16:43');
COMMIT;

SELECT *,
	COALESCE(
		(SELECT SUM(pt.rupees) FROM (
			SELECT  MONTH(`date`) `month`,
			MAX(id) id,
			SUM(rupees) rupees
			FROM house_details 
			GROUP BY `month`,subincome
		) pt 
		WHERE CASE WHEN pt.month = t1.month THEN pt.id < t1.id ELSE pt.month < t1.month END 
	), 0) AS progressive_total,
	(SELECT SUM(rupees) FROM(
			SELECT  MONTH(`date`) `month`,
			MAX(id) id,
			SUM(rupees) rupees
			FROM house_details 
			GROUP BY `month`,subincome
		) cs 
		WHERE  CASE WHEN cs.month = t1.month THEN cs.id <= t1.id ELSE cs.month <= t1.month END 
	) AS cumulative_sum 
	FROM (
		SELECT MONTHNAME(t.date) AS `monthname`,
		MAX(id) id,
		MONTH(t.date) `month`,
		YEAR(t.date) AS `year`,
		GROUP_CONCAT(t.income) income,
		t.subincome,
		GROUP_CONCAT(t.ssubincome) ssubincome,
		SUM(rupees) AS amount,
		GROUP_CONCAT(receipt_id) AS receipt_ids 
		FROM house_details t 
		WHERE YEAR(t.date) = YEAR(CURRENT_DATE()) 
		GROUP BY `monthname`,`month`, t.subincome
		ORDER BY `month`
	) t1


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